The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 1, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 1, b(n) = n Then the exact value of C equals exp(-1) - ------------- 5 exp(-1) - 2 Its floating-point approximation is 2.29061669278536242210753341456184502578206873869073466505713149509941880304\ 8701082501193994871043746 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) n 2 (n + 4) (-1) P(n) = 1/2 (n + 7 n + 11) exp(-1) G[n] + ------------- 2 (n + 1) 2 (5 n + 20) (-1) Q(n) = -5/2 (n + 7 n + 11) exp(-1) G[n] + ---------------------- 2 2 + (n + 1)! (n + 7 n + 11) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 4) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 4) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 2, P(1) = 7 Q(0) = 1, Q(1) = 3 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ------------- 5 exp(-1) - 2 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -97 equals, 0.11 10 This ends this article, that took, 1.069, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 2, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 2, b(n) = n Then the exact value of C equals exp(-1) - -------------- 16 exp(-1) - 6 Its floating-point approximation is 3.22902536539711983767523123810332605392203242787676723636312613044916575983\ 8868849052249466203800033 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) n 3 2 (n + 3) (n + 6) (-1) P(n) = 1/6 (n + 12 n + 44 n + 49) exp(-1) G[n] + --------------------- 6 3 2 Q(n) = -8/3 (n + 12 n + 44 n + 49) exp(-1) G[n] 2 (n + 1) (8 n + 72 n + 144) (-1) 3 2 + ------------------------------- + (n + 1)! (n + 12 n + 44 n + 49) 3 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 5) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 5) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 3, P(1) = 13 Q(0) = 1, Q(1) = 4 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - -------------- 16 exp(-1) - 6 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -97 equals, 0.28 10 This ends this article, that took, 0.792, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 3, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 3, b(n) = n Then the exact value of C equals exp(-1) - --------------- 65 exp(-1) - 24 Its floating-point approximation is 4.18823813452741189832266056878199448516214473532582205205970830738243139863\ 7242901726134963076872277 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 4 3 2 P(n) = 1/24 (n + 18 n + 113 n + 292 n + 261) exp(-1) G[n] 3 2 n (n + 15 n + 69 n + 96) (-1) + ------------------------------ 24 65 4 3 2 Q(n) = - -- (n + 18 n + 113 n + 292 n + 261) exp(-1) G[n] 24 3 2 (n + 1) (65 n + 975 n + 4485 n + 6240) (-1) + -------------------------------------------- 24 4 3 2 + (n + 1)! (n + 18 n + 113 n + 292 n + 261) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 6) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 6) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 4, P(1) = 21 Q(0) = 1, Q(1) = 5 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - --------------- 65 exp(-1) - 24 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -96 equals, 0.169 10 This ends this article, that took, 0.778, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 4, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 4, b(n) = n Then the exact value of C equals exp(-1) - ----------------- 326 exp(-1) - 120 Its floating-point approximation is 5.15944184208397258530685064817159502170169272321287479996217023631666195524\ 7122874173202682687408593 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 5 4 3 2 P(n) = 1/120 (n + 25 n + 235 n + 1035 n + 2129 n + 1631) exp(-1) G[n] 4 3 2 n (n + 22 n + 170 n + 543 n + 600) (-1) + ----------------------------------------- 120 163 5 4 3 2 Q(n) = - --- (n + 25 n + 235 n + 1035 n + 2129 n + 1631) exp(-1) G[n] 60 4 3 2 (n + 1) (163 n + 3586 n + 27710 n + 88509 n + 97800) (-1) + ----------------------------------------------------------- 60 5 4 3 2 + (n + 1)! (n + 25 n + 235 n + 1035 n + 2129 n + 1631) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 7) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 7) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 5, P(1) = 31 Q(0) = 1, Q(1) = 6 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ----------------- 326 exp(-1) - 120 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -95 equals, 0.4443 10 This ends this article, that took, 0.811, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 5, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 5, b(n) = n Then the exact value of C equals exp(-1) - ------------------ 1957 exp(-1) - 720 Its floating-point approximation is 6.13811405373262214095017099829807014206414253909145820552878284235697513074\ 0786595004183278820168330 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 1/720 6 5 4 3 2 (n + 33 n + 430 n + 2825 n + 9844 n + 17203 n + 11743) exp(-1) G[n] 5 4 3 2 n (n + 30 n + 341 n + 1828 n + 4599 n + 4320) (-1) + ----------------------------------------------------- 720 1957 Q(n) = - ---- 720 6 5 4 3 2 (n + 33 n + 430 n + 2825 n + 9844 n + 17203 n + 11743) exp(-1) G[n] + 5 4 3 2 (1957 n + 58710 n + 667337 n + 3577396 n + 9000243 n + 8454240) (n + 1) (-1) /720 6 5 4 3 2 + (n + 1)! (n + 33 n + 430 n + 2825 n + 9844 n + 17203 n + 11743) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 8) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 8) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 6, P(1) = 43 Q(0) = 1, Q(1) = 7 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ------------------ 1957 exp(-1) - 720 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -94 equals, 0.11488 10 This ends this article, that took, 0.702, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 6, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 6, b(n) = n Then the exact value of C equals exp(-1) - -------------------- 13700 exp(-1) - 5040 Its floating-point approximation is 7.12172426098293840297208800555302246098471280732706716619388718715020128990\ 9485349542480868135216986 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 1/5040 7 6 5 4 3 2 (n + 42 n + 721 n + 6545 n + 33859 n + 99589 n + 153796 n + 95901) exp(-1) G[n] 6 5 4 3 2 n (n + 39 n + 605 n + 4765 n + 20031 n + 42453 n + 35280) (-1) + ------------------------------------------------------------------ 5040 685 Q(n) = - --- 252 7 6 5 4 3 2 (n + 42 n + 721 n + 6545 n + 33859 n + 99589 n + 153796 n + 95901) 6 5 4 3 2 exp(-1) G[n] + (685 n + 26715 n + 414425 n + 3264025 n + 13721235 n (n + 1) + 29080305 n + 24166800) (-1) /252 + (n + 1)! 7 6 5 4 3 2 (n + 42 n + 721 n + 6545 n + 33859 n + 99589 n + 153796 n + 95901) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 9) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 9) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 7, P(1) = 57 Q(0) = 1, Q(1) = 8 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - -------------------- 13700 exp(-1) - 5040 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -94 equals, 0.45331 10 This ends this article, that took, 0.830, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 7, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 7, b(n) = n Then the exact value of C equals exp(-1) - ---------------------- 109601 exp(-1) - 40320 Its floating-point approximation is 8.10875667470144027761985164540739497446014256733944970225775092415774174527\ 7137945133972020090622099 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 8 7 6 5 4 3 P(n) = 1/40320 (n + 52 n + 1134 n + 13524 n + 96299 n + 418404 n 2 + 1080778 n + 1513440 n + 876809) exp(-1) G[n] + 7 6 5 4 3 2 (n + 49 n + 988 n + 10605 n + 65294 n + 229967 n + 427617 n + 322560) n (-1) /40320 109601 8 7 6 5 4 3 Q(n) = - ------ (n + 52 n + 1134 n + 13524 n + 96299 n + 418404 n 40320 2 7 6 + 1080778 n + 1513440 n + 876809) exp(-1) G[n] + (109601 n + 5370449 n 5 4 3 2 + 108285788 n + 1162318605 n + 7156287694 n + 25204613167 n (n + 1) 8 7 + 46867250817 n + 35352898560) (-1) /40320 + (n + 1)! (n + 52 n 6 5 4 3 2 + 1134 n + 13524 n + 96299 n + 418404 n + 1080778 n + 1513440 n + 876809) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 10) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 10) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 8, P(1) = 73 Q(0) = 1, Q(1) = 9 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ---------------------- 109601 exp(-1) - 40320 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -93 equals, 0.763920 10 This ends this article, that took, 0.727, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 8, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 8, b(n) = n Then the exact value of C equals exp(-1) - ----------------------- 986410 exp(-1) - 362880 Its floating-point approximation is 9.09825234537949395945132748748340209978693738225603835578695324387533516509\ 1986283586994472742898269 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 9 8 7 6 5 4 P(n) = 1/362880 (n + 63 n + 1698 n + 25662 n + 239295 n + 1425333 n 3 2 8 + 5412350 n + 12605286 n + 16294281 n + 8877691) exp(-1) G[n] + (n 7 6 5 4 3 2 + 60 n + 1519 n + 21161 n + 177109 n + 910081 n + 2797006 n n + 4686723 n + 3265920) (-1) /362880 98641 9 8 7 6 5 4 Q(n) = - ----- (n + 63 n + 1698 n + 25662 n + 239295 n + 1425333 n 36288 3 2 + 5412350 n + 12605286 n + 16294281 n + 8877691) exp(-1) G[n] + ( 8 7 6 5 4 98641 n + 5918460 n + 149835679 n + 2087342201 n + 17470208869 n 3 2 + 89771299921 n + 275899468846 n + 462303043443 n + 322153614720) (n + 1) 9 8 7 6 5 (-1) /36288 + (n + 1)! (n + 63 n + 1698 n + 25662 n + 239295 n 4 3 2 + 1425333 n + 5412350 n + 12605286 n + 16294281 n + 8877691) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 11) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 11) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 9, P(1) = 91 Q(0) = 1, Q(1) = 10 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ----------------------- 986410 exp(-1) - 362880 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -91 equals, 0.19826174 10 This ends this article, that took, 0.870, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 9, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 9, b(n) = n Then the exact value of C equals exp(-1) - ------------------------- 9864101 exp(-1) - 3628800 Its floating-point approximation is 10.0895769440159259001618983731093260891832388557844282551560904085209349949\ 8181773793998786112949971 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 10 9 8 7 6 5 P(n) = 1/3628800 (n + 75 n + 2445 n + 45570 n + 537033 n + 4175157 n 4 3 2 + 21649655 n + 73787850 n + 157830711 n + 190788103 n + 98641011) 9 8 7 6 5 4 exp(-1) G[n] + (n + 72 n + 2230 n + 38948 n + 422149 n + 2939944 n 3 2 n + 13130332 n + 36181092 n + 55658043 n + 36288000) (-1) /3628800 9864101 10 9 8 7 6 5 Q(n) = - ------- (n + 75 n + 2445 n + 45570 n + 537033 n + 4175157 n 3628800 4 3 2 + 21649655 n + 73787850 n + 157830711 n + 190788103 n + 98641011) 9 8 7 exp(-1) G[n] + (9864101 n + 710215272 n + 21996945230 n 6 5 4 + 384187005748 n + 4164120373049 n + 28999904550344 n 3 2 + 129518921011532 n + 356893945778292 n + 549016557614343 n (n + 1) 10 9 8 + 357948497088000) (-1) /3628800 + (n + 1)! (n + 75 n + 2445 n 7 6 5 4 3 + 45570 n + 537033 n + 4175157 n + 21649655 n + 73787850 n 2 + 157830711 n + 190788103 n + 98641011) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 12) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 12) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 10, P(1) = 111 Q(0) = 1, Q(1) = 11 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ------------------------- 9864101 exp(-1) - 3628800 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -90 equals, 0.22158356 10 This ends this article, that took, 0.948, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 10, and , b(n) = n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 10, b(n) = n Then the exact value of C equals exp(-1) - ---------------------------- 108505112 exp(-1) - 39916800 Its floating-point approximation is 11.0822950689777142737421360898031943817332211776552189208171741405912865813\ 6754204773026199990047457 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, -1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / -1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence G[n + 1] = exp(-1) + (n + 2) G[n] and hence, iterating, we also have 2 G[n + 2] = (n + 5 n + 6) G[n] + exp(-1) n + 4 exp(-1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 11 10 9 8 7 6 P(n) = 1/39916800 (n + 88 n + 3410 n + 76725 n + 1112463 n + 10899966 n 5 4 3 2 + 73533746 n + 340980035 n + 1062949261 n + 2116534486 n 10 9 8 + 2415943540 n + 1193556233) exp(-1) G[n] + (n + 85 n + 3156 n 7 6 5 4 3 + 67338 n + 913283 n + 8216271 n + 49578514 n + 197774598 n 2 n + 498042179 n + 712975329 n + 439084800) (-1) /39916800 13563139 11 10 9 8 7 6 Q(n) = - -------- (n + 88 n + 3410 n + 76725 n + 1112463 n + 10899966 n 4989600 5 4 3 2 + 73533746 n + 340980035 n + 1062949261 n + 2116534486 n 10 9 + 2415943540 n + 1193556233) exp(-1) G[n] + (13563139 n + 1152866815 n 8 7 6 + 42805266684 n + 913314653982 n + 12386984275337 n 5 4 3 + 111438425634669 n + 672440276795446 n + 2682444363343122 n 2 (n + 1) + 6755015301639881 n + 9670183490797731 n + 5955368175187200) (-1) 11 10 9 8 7 /4989600 + (n + 1)! (n + 88 n + 3410 n + 76725 n + 1112463 n 6 5 4 3 2 + 10899966 n + 73533746 n + 340980035 n + 1062949261 n + 2116534486 n + 2415943540 n + 1193556233) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 13) P(n + 1) - (n + 2) P(n) = 0 Q(n + 2) - (n + 13) Q(n + 1) - (n + 2) Q(n) = 0 subject to the initial conditions P(0) = 11, P(1) = 133 Q(0) = 1, Q(1) = 12 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(-1) - ---------------------------- 108505112 exp(-1) - 39916800 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -89 equals, 0.257997357 10 This ends this article, that took, 0.772, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 1, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 1, b(n) = -n Then the exact value of C equals exp(1) ----------- -1 + exp(1) Its floating-point approximation is 1.58197670686932642438500200510901155854686930107539613626678705964804381739\ 1669743287204709404875058 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = exp(1) G[n] Q(n) = -(n + 1)! + exp(1) G[n] These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 4) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 4) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 2, P(1) = 5 Q(0) = 1, Q(1) = 3 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) ----------- -1 + exp(1) QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ equals, 0. This ends this article, that took, 0.228, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 2, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 2, b(n) = -n Then the exact value of C equals exp(1) Its floating-point approximation is 2.71828182845904523536028747135266249775724709369995957496696762772407663035\ 3547594571382178525166427 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 1 + (n + 1) exp(1) G[n] Q(n) = (n + 1)! (n + 1) These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 5) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 5) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 3, P(1) = 11 Q(0) = 1, Q(1) = 4 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ equals, 0. This ends this article, that took, 0.507, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 3, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 3, b(n) = -n Then the exact value of C equals exp(1) ---------- exp(1) - 2 Its floating-point approximation is 3.78442238235466562875310575695963305674795677063057424718264913416655914092\ 3221853383421174535225999 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 2 P(n) = 1/2 (n + 3 n + 3) exp(1) G[n] + n/2 + 1 Q(n) = 2 2 1/2 (n + 3 n + 3) exp(1) G[n] + 1/2 (-2 n - 6 n - 6) (n + 1)! + 1 + n/2 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 6) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 6) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 4, P(1) = 19 Q(0) = 1, Q(1) = 5 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) ---------- exp(1) - 2 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -98 equals, -0.1 10 This ends this article, that took, 0.689, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 4, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 4, b(n) = -n Then the exact value of C equals exp(1) - ------------ 2 exp(1) - 6 Its floating-point approximation is 4.82447016745576732333958902719654340072963792503094683880112203402605190925\ 7686025226450837167052667 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 2 3 2 n 5 n P(n) = 1/6 (n + 6 n + 14 n + 11) exp(1) G[n] + ---- + --- + 4/3 6 6 3 2 Q(n) = -1/3 (n + 6 n + 14 n + 11) exp(1) G[n] 2 3 2 n 5 n + 1/3 (3 n + 18 n + 42 n + 33) (n + 1)! - ---- - --- - 8/3 3 3 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 7) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 7) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 5, P(1) = 29 Q(0) = 1, Q(1) = 6 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) - ------------ 2 exp(1) - 6 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -98 equals, 0.8 10 This ends this article, that took, 0.819, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 5, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 5, b(n) = -n Then the exact value of C equals exp(1) ------------- 9 exp(1) - 24 Its floating-point approximation is 5.85160064959487971142263331218369076252805764955869730909000200968740579526\ 9100176092887939917866860 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 3 2 4 3 2 n 3 n 31 n 19 1/24 (n + 10 n + 41 n + 76 n + 53) exp(1) G[n] + ---- + ---- + ---- + -- 24 8 24 12 4 3 2 Q(n) = 3/8 (n + 10 n + 41 n + 76 n + 53) exp(1) G[n] 3 2 4 3 2 3 n 27 n + 1/8 (-8 n - 80 n - 328 n - 608 n - 424) (n + 1)! + ---- + ----- 8 8 93 n + ---- + 57/4 8 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 8) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 8) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 6, P(1) = 41 Q(0) = 1, Q(1) = 7 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) ------------- 9 exp(1) - 24 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -97 equals, -0.86 10 This ends this article, that took, 0.634, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 6, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 6, b(n) = -n Then the exact value of C equals exp(1) - --------------- 44 exp(1) - 120 Its floating-point approximation is 6.87129660173296177453289854266958482649469973997463787591577677301215234417\ 5097916975436723383126180 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 4 5 4 3 2 n P(n) = 1/120 (n + 15 n + 95 n + 305 n + 489 n + 309) exp(1) G[n] + --- 120 3 2 7 n 2 n 71 n 37 + ---- + ---- + ---- + -- 60 3 40 20 11 5 4 3 2 Q(n) = - -- (n + 15 n + 95 n + 305 n + 489 n + 309) exp(1) G[n] 30 5 4 3 2 + 1/30 (30 n + 450 n + 2850 n + 9150 n + 14670 n + 9270) (n + 1)! 4 3 2 11 n 77 n 88 n 781 n - ----- - ----- - ----- - ----- - 407/5 30 15 3 10 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 9) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 9) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 7, P(1) = 55 Q(0) = 1, Q(1) = 8 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) - --------------- 44 exp(1) - 120 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -96 equals, 0.119 10 This ends this article, that took, 0.653, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 7, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 7, b(n) = -n Then the exact value of C equals exp(1) ---------------- 265 exp(1) - 720 Its floating-point approximation is 7.88628876557801878356687399657724766279779410550411850293645104376981001456\ 7661380120532454441256414 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 6 5 4 3 2 1/720 (n + 21 n + 190 n + 925 n + 2524 n + 3623 n + 2119) exp(1) G[n] 5 4 3 2 n n 169 n 41 n 331 n 761 + --- + ---- + ------ + ----- + ----- + --- 720 36 720 40 144 360 53 6 5 4 3 2 Q(n) = --- (n + 21 n + 190 n + 925 n + 2524 n + 3623 n + 2119) exp(1) G[n] 144 + 1/144 6 5 4 3 2 (-144 n - 3024 n - 27360 n - 133200 n - 363456 n - 521712 n - 305136) 5 4 3 2 53 n 265 n 8957 n 2173 n 87715 n 40333 (n + 1)! + ----- + ------ + ------- + ------- + ------- + ----- 144 36 144 8 144 72 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 10) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 10) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 8, P(1) = 71 Q(0) = 1, Q(1) = 9 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) ---------------- 265 exp(1) - 720 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -95 equals, -0.1561 10 This ends this article, that took, 0.870, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 8, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 8, b(n) = -n Then the exact value of C equals exp(1) - ------------------ 1854 exp(1) - 5040 Its floating-point approximation is 8.89810304707489785707025925250519881901366512271426655044069973032160045776\ 0197655691438116523114082 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) P(n) = 1/5040 7 6 5 4 3 2 (n + 28 n + 343 n + 2345 n + 9569 n + 23121 n + 30414 n + 16687) 6 5 4 3 2 n 3 n n 401 n 1043 n 137 n 5993 exp(1) G[n] + ---- + ---- + ---- + ------ + ------- + ----- + ---- 5040 560 16 1008 720 48 2520 103 Q(n) = - --- 280 7 6 5 4 3 2 (n + 28 n + 343 n + 2345 n + 9569 n + 23121 n + 30414 n + 16687) 7 6 5 4 3 exp(1) G[n] + 1/280 (280 n + 7840 n + 96040 n + 656600 n + 2679320 n 6 5 4 2 103 n 2781 n 927 n + 6473880 n + 8515920 n + 4672360) (n + 1)! - ------ - ------- - ------ 280 280 8 3 2 41303 n 107429 n 42333 n 617279 - -------- - --------- - ------- - ------ 56 40 8 140 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 11) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 11) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 9, P(1) = 89 Q(0) = 1, Q(1) = 10 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) - ------------------ 1854 exp(1) - 5040 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -94 equals, 0.13124 10 This ends this article, that took, 0.673, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 9, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 9, b(n) = -n Then the exact value of C equals exp(1) -------------------- 14833 exp(1) - 40320 Its floating-point approximation is 9.90766379877657298491417952126842395690873204525313022535022548412267125460\ 2703472413851496306014338 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 8 7 6 5 4 3 P(n) = 1/40320 (n + 36 n + 574 n + 5236 n + 29659 n + 106148 n 7 6 5 2 n n 269 n + 233050 n + 285360 n + 148329) exp(1) G[n] + ----- + ---- + ------ 40320 1152 20160 4 3 2 311 n 6131 n 11137 n 46217 n 5919 + ------ + ------- + -------- + ------- + ---- 2688 10080 5760 13440 2240 2119 8 7 6 5 4 3 2 Q(n) = ---- (n + 36 n + 574 n + 5236 n + 29659 n + 106148 n + 233050 n 5760 8 7 + 285360 n + 148329) exp(1) G[n] + 1/5760 (-5760 n - 207360 n 6 5 4 3 2 - 3306240 n - 30159360 n - 170835840 n - 611412480 n - 1342368000 n 7 6 5 2119 n 14833 n 570011 n - 1643673600 n - 854375040) (n + 1)! + ------- + -------- + --------- 5760 1152 2880 4 3 2 659009 n 12991589 n 165195121 n 97933823 n 12542361 + --------- + ----------- + ------------ + ---------- + -------- 384 1440 5760 1920 320 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 12) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 12) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 10, P(1) = 109 Q(0) = 1, Q(1) = 11 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) -------------------- 14833 exp(1) - 40320 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -93 equals, -0.290033 10 This ends this article, that took, 0.896, seconds to generate. The exact value of the infinite generalized continued fraction a(1)+b(1)/(a(\ 2)+b(2)/... where , a(n) = n + 10, and , b(n) = -n By Shalosh B. Ekhad Theorem: Consider the following Infinite continued fraction, let's call it C\ : C=a(1)+ b(1)/(a(2)+ b(2)/(a(3)+b(3)/(a(4)+.. where a(n) = n + 10, b(n) = -n Then the exact value of C equals exp(1) - ---------------------- 133496 exp(1) - 362880 Its floating-point approximation is 10.9155656666388710295854452486713966433151243231907215601811402789349909743\ 6949346826013381716603749 Proof: Let's denote the incomplete Gamma function G[n] = GAMMA(n + 2, 1) in other words infinity / | (n + 1) G[n] = | x exp(-x) dx | / 1 By integration by parts it is very easy to deduce that G[n] satisfies the fi\ rst-order (inhomog.) linear recurrence (n + 2) G[n + 1] = (-1) exp(1) + (n + 2) G[n] and hence, iterating, we also have 2 n n G[n + 2] = (n + 5 n + 6) G[n] + n (-1) exp(1) + 2 (-1) exp(1) Let P(n) and Q(n) be the numerator and denominator of the truncated version \ of the above infinite continued fraction truncated at a(n),b(n) Then we have the following explicit expressions for P(n) and Q(n) 9 8 7 6 5 4 P(n) = 1/362880 (n + 45 n + 906 n + 10626 n + 79527 n + 391839 n 8 3 2 n + 1264934 n + 2567778 n + 2959881 n + 1468457) exp(1) G[n] + ------ 362880 7 6 5 4 3 2 11 n 41 n 463 n 69029 n 20981 n 89869 n 489827 n + ----- + ----- + ------ + -------- + -------- + -------- + -------- 90720 17280 17280 362880 24192 36288 120960 527383 + ------ 181440 16687 9 8 7 6 5 4 Q(n) = - ----- (n + 45 n + 906 n + 10626 n + 79527 n + 391839 n 45360 3 2 + 1264934 n + 2567778 n + 2959881 n + 1468457) exp(1) G[n] + 1/45360 ( 9 8 7 6 5 45360 n + 2041200 n + 41096160 n + 481995360 n + 3607344720 n 4 3 2 + 17773817040 n + 57377406240 n + 116474410080 n + 134260202160 n 8 7 6 5 16687 n 183557 n 684167 n 7726081 n + 66609209520) (n + 1)! - -------- - --------- - --------- - ---------- 45360 11340 2160 2160 4 3 2 1151886923 n 350109947 n 1499644003 n 8173743149 n 8800440121 - ------------- - ------------ - ------------- - ------------ - ---------- 45360 3024 4536 15120 22680 These follow from the fact that by the general theory of continued fractions\ (and is also easily seen) P(n) and Q(n) both satisfy the following second-order linear recurrence P(n + 2) - (n + 13) P(n + 1) - (-n - 2) P(n) = 0 Q(n + 2) - (n + 13) Q(n + 1) - (-n - 2) Q(n) = 0 subject to the initial conditions P(0) = 11, P(1) = 131 Q(0) = 1, Q(1) = 12 But the above proposed expressions for P(n) and Q(n) in terms of G[n] and n \ satisfy the correct initial conditions and satisfy the the recurrence, as is readily checked by using the above expressions of G[n+\ 1] and G[n+2] in terms of G[n] and simplifying Finally, taking the limit as n goes to infinity of P(n)/Q(n) we get that C indeed equals exp(1) - ---------------------- 133496 exp(1) - 362880 QED. Comment: Just as a sanity check, the difference between C and P(100)/Q(100) \ -92 equals, 0.420167 10 This ends this article, that took, 0.702, seconds to generate. -------------- This took, 14.997, seconds.