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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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 floatingpoint 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\
rstorder (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 secondorder 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.