------------------------------ The input for log(2) is: RAsumV(binomial(n,k)*binomial(n+k,k),k,n,10); the output is A Proof of the Irrationality of, ln(2) By Shalosh B. Ekhad Let n ----- \ b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k = 0 Thanks to the Zeilberger algorithm, we know that , b(n), satisfies the following recurrence (-n - 1) b(n) + (6 n + 9) b(n + 1) + (-n - 2) b(n + 2) = 0 subject to the initial conditons:, b(0) = 1, b(1) = 3 Let , a(n), be the sequence of rational numbers satisfying the same recurrence as, b(n), i.e. (-n - 1) a(n) + (6 n + 9) a(n + 1) + (-n - 2) a(n + 2) = 0 but subject to the intial conditions: a(0) = 0, a(1) = 2 a(n) It can be seen that, ----, tends to , ln(2), as n goes to infinity. b(n) We now need a lemma, whose proof is left to the reader Crucial Lemma: Let L(m)=lcm(1,2, ..., m), then a(n) L(n), are always integers. Consider c(n) = a(n) b(n - 1) - a(n - 1) b(n) It is readily seen that , c(n), satisfies the recurrence n c(n) - ------ + c(n + 1) = 0 n + 1 subject to the intial conditions:, c(1) = 2 n By the Poincare Lemma, , b(n) = O(alpha ) 1/2 where alpha=, 3 + 2 2 , that equals , 5.828427125, is the highest root of the equation, in N 2 -N + 6 N - 1 = 0 n Also by the Poincare Lemma, , c(n) = O(beta ) where beta=, 1, that equals, 1., is the absolute value of the highest root of the equation, in N N - 1 = 0 Since a(n) a(n - 1) c(n) ---- - -------- = ------------- b(n) b(n - 1) b(n) b(n - 1) We have n ----- a(n) \ c(i) ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = 1 and hence infinity ----- a(n) \ c(i) ln(2) - ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = n + 1 and hence | a(n) | / beta \n | -ln(2) + ---- | = O(|------| ) | b(n) | | 2| \alpha / i.e. : | a(n) | 1 | -ln(2) + ---- | = O(------------) | b(n) | n 33.97056275 Let A(n) = a(n) L(n) B(n) = b(n) L(n) By the lemma A(n) are integers, and of course B(n) are integers , and of cou\ rse a(n)/b(n)=A(n)/B(n), and | A(n) | / beta \n | -ln(2) + ---- | = O(|------| ) | B(n) | | 2| \alpha / By the prime number theory n L(n) = O(e ) So we have n B(n) = O((alpha exp(1)) ) It follows that | A(n) | 1 | -ln(2) + ---- | = O(---------------) | B(n) | (1 + delta) B(n) 1/2 ln(3 + 2 2 ) - 1 where delta=, ------------------, that equals, 0.276082872, 1/2 ln(3 + 2 2 ) + 1 is positive!. Since A(n), B(n) are integers it follows that , ln(2), is irrational! 1/2 1 2 ln(3 + 2 2 ) Furthermore it has an irrationality measure, 1 + -----, = , ------------------, delta 1/2 ln(3 + 2 2 ) - 1 that equals, 4.622100832 This finishes this nice article, that took, 0.672, seconds to produce. Have a nice day. -------------------- ------------------------------ The input for Zeta(2) is: RAsumV(binomial(n,k)^2*binomial(n+k,k),k,n,10); the output is 2 A Proof of the Irrationality of, Pi By Shalosh B. Ekhad Let n ----- \ 2 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k = 0 Thanks to the Zeilberger algorithm, we know that , b(n), satisfies the following recurrence 2 2 2 -(n + 1) b(n) + (-11 n - 33 n - 25) b(n + 1) + (n + 2) b(n + 2) = 0 subject to the initial conditons:, b(0) = 1, b(1) = 3 Let , a(n), be the sequence of rational numbers satisfying the same recurrence as, b(n), i.e. 2 2 2 -(n + 1) a(n) + (-11 n - 33 n - 25) a(n + 1) + (n + 2) a(n + 2) = 0 but subject to the intial conditions: a(0) = 0, a(1) = 30 a(n) 2 It can be seen that, ----, tends to , Pi , as n goes to infinity. b(n) We now need a lemma, whose proof is left to the reader Crucial Lemma: Let L(m)=lcm(1,2, ..., m), then 2 a(n) L(n) , are always integers. Consider c(n) = a(n) b(n - 1) - a(n - 1) b(n) It is readily seen that , c(n), satisfies the recurrence 2 n c(n) -------- + c(n + 1) = 0 2 (n + 1) subject to the intial conditions:, c(1) = 30 n By the Poincare Lemma, , b(n) = O(alpha ) 1/2 5 5 where alpha=, 11/2 + ------, that equals , 11.09016994, 2 is the highest root of the equation, in N 2 N - 11 N - 1 = 0 n Also by the Poincare Lemma, , c(n) = O(beta ) where beta=, 1, that equals, 1., is the absolute value of the highest root of the equation, in N N + 1 = 0 Since a(n) a(n - 1) c(n) ---- - -------- = ------------- b(n) b(n - 1) b(n) b(n - 1) We have n ----- a(n) \ c(i) ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = 1 and hence infinity ----- 2 a(n) \ c(i) Pi - ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = n + 1 and hence | 2 a(n) | / beta \n | -Pi + ---- | = O(|------| ) | b(n) | | 2| \alpha / i.e. : | 2 a(n) | 1 | -Pi + ---- | = O(------------) | b(n) | n 122.9918693 Let 2 A(n) = a(n) L(n) 2 B(n) = b(n) L(n) By the lemma A(n) are integers, and of course B(n) are integers , and of cou\ rse a(n)/b(n)=A(n)/B(n), and | 2 A(n) | / beta \n | -Pi + ---- | = O(|------| ) | B(n) | | 2| \alpha / By the prime number theory n L(n) = O(e ) So we have n B(n) = O((alpha exp(2)) ) It follows that | 2 A(n) | 1 | -Pi + ---- | = O(---------------) | B(n) | (1 + delta) B(n) 1/2 5 5 ln(11/2 + ------) - 2 2 where delta=, ---------------------, that equals, 0.092159255, 1/2 5 5 ln(11/2 + ------) + 2 2 2 is positive!. Since A(n), B(n) are integers it follows that , Pi , is irrational! 1 Furthermore it has an irrationality measure, 1 + -----, = , delta 1/2 5 5 2 ln(11/2 + ------) 2 ---------------------, that equals, 11.85078220 1/2 5 5 ln(11/2 + ------) - 2 2 This finishes this nice article, that took, 1.039, seconds to produce. Have a nice day. -------------------- ------------------------------ The input for Zeta(3) is: RAsumV(binomial(n,k)^2*binomial(n+k,k)^2,k,n,10); the output is A Proof of the Irrationality of, Zeta(3) By Shalosh B. Ekhad Let n ----- \ 2 2 b(n) = ) binomial(n, k) binomial(n + k, k) / ----- k = 0 Thanks to the Zeilberger algorithm, we know that , b(n), satisfies the following recurrence 3 2 3 (n + 1) b(n) - (17 n + 51 n + 39) (2 n + 3) b(n + 1) + (n + 2) b(n + 2) = 0 subject to the initial conditons:, b(0) = 1, b(1) = 5 Let , a(n), be the sequence of rational numbers satisfying the same recurrence as, b(n), i.e. 3 2 3 (n + 1) a(n) - (17 n + 51 n + 39) (2 n + 3) a(n + 1) + (n + 2) a(n + 2) = 0 but subject to the intial conditions: a(0) = 0, a(1) = 6 a(n) It can be seen that, ----, tends to , Zeta(3), as n goes to infinity. b(n) We now need a lemma, whose proof is left to the reader Crucial Lemma: Let L(m)=lcm(1,2, ..., m), then 3 a(n) L(n) , are always integers. Consider c(n) = a(n) b(n - 1) - a(n - 1) b(n) It is readily seen that , c(n), satisfies the recurrence 3 n c(n) - -------- + c(n + 1) = 0 3 (n + 1) subject to the intial conditions:, c(1) = 6 n By the Poincare Lemma, , b(n) = O(alpha ) 1/2 where alpha=, 17 + 12 2 , that equals , 33.97056275, is the highest root of the equation, in N 2 N - 34 N + 1 = 0 n Also by the Poincare Lemma, , c(n) = O(beta ) where beta=, 1, that equals, 1., is the absolute value of the highest root of the equation, in N N - 1 = 0 Since a(n) a(n - 1) c(n) ---- - -------- = ------------- b(n) b(n - 1) b(n) b(n - 1) We have n ----- a(n) \ c(i) ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = 1 and hence infinity ----- a(n) \ c(i) Zeta(3) - ---- = ) ------------- b(n) / b(i - 1) b(i) ----- i = n + 1 and hence | a(n) | / beta \n | -Zeta(3) + ---- | = O(|------| ) | b(n) | | 2| \alpha / i.e. : | a(n) | 1 | -Zeta(3) + ---- | = O(------------) | b(n) | n 1153.999134 Let 3 A(n) = a(n) L(n) 3 B(n) = b(n) L(n) By the lemma A(n) are integers, and of course B(n) are integers , and of cou\ rse a(n)/b(n)=A(n)/B(n), and | A(n) | / beta \n | -Zeta(3) + ---- | = O(|------| ) | B(n) | | 2| \alpha / By the prime number theory n L(n) = O(e ) So we have n B(n) = O((alpha exp(3)) ) It follows that | A(n) | 1 | -Zeta(3) + ---- | = O(---------------) | B(n) | (1 + delta) B(n) 1/2 ln(17 + 12 2 ) - 3 where delta=, --------------------, that equals, 0.080529431, 1/2 ln(17 + 12 2 ) + 3 is positive!. Since A(n), B(n) are integers it follows that , Zeta(3), is irrational! 1 Furthermore it has an irrationality measure, 1 + -----, = , delta 1/2 2 ln(17 + 12 2 ) --------------------, that equals, 13.41782024 1/2 ln(17 + 12 2 ) - 3 This finishes this nice article, that took, 1.798, seconds to produce. Have a nice day. -------------------- -------------------- the whole thing took, 3.560, seconds.