Question

Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$

score 135 · Answer 1 · 2026-03-28 16:05:19

Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$

score 166 · Answer 2 · 2026-03-28 16:05:21

166

Views

How does one show that $\prod_{k=1}^{\infty}\left(\prod_{j=1}^{\infty}{(2j)^{2^k}-1\over (2j)^{2^k}+1}\right)^{1/2^k}={2\over \pi}?$

Published on 28 Mar 2026 - 16:05

score 219 · Answer 3 · 2026-03-25 06:10:33

219

Views

Parity of Ramanujan's Tau Function

Published on 25 Mar 2026 - 6:10

score 83 · Answer 4 · 2017-03-24 06:55:17

83

Views

How can I be certain that this infinite product has no associated value?

Published on 24 Mar 2017 - 6:55

score 918 · Answer 5 · 2017-03-31 08:09:49

918

Views

Find $\prod^\infty_{n=1} (1 - 10^{-3* 2^n})$.

Published on 31 Mar 2017 - 8:09

score 45 · Answer 6 · 2023-01-29 20:23:50

45

Views

References for the Euler function $\prod_{k=1}^\infty{(1-q^k)}$

Published on 29 Jan 2023 - 20:23

score 86 · Answer 7 · 2023-01-31 04:46:06

86

Views

Limit of $\frac{1}{n!}\prod_{i=1}^{n}(i-\alpha)$

Published on 31 Jan 2023 - 4:46

score 239 · Answer 8 · 2023-02-04 13:45:02

239

Views

Evaluating the limit $\lim_{n \to \infty} \left[\log(n) + 2n \log(4n + 2) - \sum_{k = 1}^{2n} (-1)^k (2k+1)\log(2k+1)\right]$

Published on 04 Feb 2023 - 13:45

score 65 · Answer 9 · 2023-02-05 09:31:14

65

Views

Closed form of a function $f(z)$ whose $\it{only}$ zeros are $n\pi$,$n\pi\omega$ and $n\pi\omega^2$

Published on 05 Feb 2023 - 9:31

score 18 · Answer 10 · 2026-03-21 01:19:40

18

Views

Showing that $\left|\prod_{k=1}^\infty \cos\left(\frac{2\pi}{3^k}\right)\right| > 0$

Published on 21 Mar 2026 - 1:19

score 153 · Answer 11 · 2023-02-09 19:21:57

153

Views

Is $(-1)^{\infty}=0$?

Published on 09 Feb 2023 - 19:21

score 325 · Answer 12 · 2023-02-11 11:03:07

325

Views

Reference for, and/or proof of, $\prod_{n=1}^\infty(\frac{4n+1}{4n-1})^{4n}(\frac{2n^2-2n+1}{2n^2+2n+1})^n=\sqrt2\cosh(\pi/2)e^{-2G/\pi}$

Published on 11 Feb 2023 - 11:03

score 54 · Answer 13 · 2023-02-13 14:43:22

54

Views

Prove that $\prod _{n=1}^{\infty } \left(1-\frac{x}{\pi n}\right) \left(\frac{x}{\pi n}+1\right)=\sin(x)/x$

Published on 13 Feb 2023 - 14:43

score 86 · Answer 14 · 2023-02-17 13:57:46

86

Views

Left Hand Limit of this continous product

Published on 17 Feb 2023 - 13:57

score 70 · Answer 15 · 2023-02-18 16:03:30

Seeking a proven explicit or closed form upper bound on the infinite product of the Zeta function over the natural numbers, 2 to infinity

List Question

Find closed form of Wallis's product type $\prod_{n=1}^{\infty}\left(\prod_{k=0}^{m}(n+k)^{{m\choose k}(-1)^k}\right)^{(-1)^n}$

How does one show that $\prod_{k=1}^{\infty}\left(\prod_{j=1}^{\infty}{(2j)^{2^k}-1\over (2j)^{2^k}+1}\right)^{1/2^k}={2\over \pi}?$

Parity of Ramanujan's Tau Function

How can I be certain that this infinite product has no associated value?

Find $\prod^\infty_{n=1} (1 - 10^{-3* 2^n})$.

References for the Euler function $\prod_{k=1}^\infty{(1-q^k)}$

Limit of $\frac{1}{n!}\prod_{i=1}^{n}(i-\alpha)$

Evaluating the limit $\lim_{n \to \infty} \left[\log(n) + 2n \log(4n + 2) - \sum_{k = 1}^{2n} (-1)^k (2k+1)\log(2k+1)\right]$

Closed form of a function $f(z)$ whose $\it{only}$ zeros are $n\pi$,$n\pi\omega$ and $n\pi\omega^2$

Showing that $\left|\prod_{k=1}^\infty \cos\left(\frac{2\pi}{3^k}\right)\right| > 0$

Is $(-1)^{\infty}=0$?

Reference for, and/or proof of, $\prod_{n=1}^\infty(\frac{4n+1}{4n-1})^{4n}(\frac{2n^2-2n+1}{2n^2+2n+1})^n=\sqrt2\cosh(\pi/2)e^{-2G/\pi}$

Prove that $\prod _{n=1}^{\infty } \left(1-\frac{x}{\pi n}\right) \left(\frac{x}{\pi n}+1\right)=\sin(x)/x$

Left Hand Limit of this continous product

Seeking a proven explicit or closed form upper bound on the infinite product of the Zeta function over the natural numbers, 2 to infinity

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