I've just worked out the limit $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^2}\right)$ that is simply solved, and the result is $\frac{1}{2}$. After that, I thought of calculating $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^3}\right)$, but I don't know how to do it. According to W|A, the result is pretty nice, but I don't see how W|A gets that. (See here.) Is there any easy way to get the answer?
Compute $\lim\limits_{n\to\infty} \prod\limits_2^n \left(1-\frac1{k^3}\right)$
-
0@Chris'ssister: Thank you very much!!! That was a great advice! In case this isn't too much to ask, when you feel like it and in case you happen to have compiled a list of those papers, books, videos, blogs,..., one would appreciate it if you can maybe post it somewhere for others to learn from you journey or email me if you can: ntbashige[at]yahoo[dot]com. Again great advice. I always neglected solving a problem from different approaches once i got an answer! – 2012-12-26
2 Answers
Since $ 1-\frac1{k^3}=\frac{(k-1)(k+\frac12+\frac{\sqrt3}2i)(k+\frac12-\frac{\sqrt3}2i)}{k^3} $ and $ k+a=\frac{\Gamma(k+a+1)}{\Gamma(k+a)}, $ every term in the product is a ratio of the Gamma functions. Also there is a formula $ \Gamma \left(\frac{1}{2}-i y\right) \Gamma \left(\frac{1}{2}+i y\right)= \pi \text{sech}\pi y. $ In particular for the end terms of the product $\frac{1}{\Gamma \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) \Gamma \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}=\frac{\cosh \frac{\sqrt{3} \pi }{2}}{\pi }. $ Multiplying those ratios and canceling out the same terms leads to a formula for the partial product: $ \prod _{k=2}^n \left(1-\frac{1}{k^3}\right)= \frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi n^3 \Gamma^2 (n)}. $ Taking the limit $n\to\infty$ gives the desired result.
The last step of Andrew getting \begin{align}\lim_{n\to \infty}\prod _{k=2}^n \left(1-\frac{1}{k^3}\right)= \frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi n^3 \Gamma^2 (n)}\end{align} was a bit ambigous.
using another method, note that \begin{align*}\Gamma(z)=\frac{1}{z e^{\gamma z}}\prod_{k=1}^{\infty}\frac{k e^{\frac{z}{k}}}{z+k} \end{align*} holds for all complex number $z$ except negative integer, we obtain
\begin{align}g(z)=\prod_{k=1}^{\infty} (1+\frac{z}{k})e^{\frac{-z}{k}}=\frac{1}{z\Gamma(z)e^{\gamma z}}\end{align}
Thus \begin{align} g(\omega)g(\omega^2)=\prod_{k=1}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}=\frac{1}{\Gamma(\omega)\Gamma(\omega^2) e^{\gamma}}=\frac{3}{e}\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}} \end{align} where $-\omega$ is the root of $x^3=1$
From \begin{align} \prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\lim_{n\to \infty}\frac{1}{n} e^{\frac{1}{2}+\cdots+\frac{1}{n}}=e^{\gamma -1} \end{align}
Thus \begin{align} \prod_{k=2}^{\infty}\left(1-\frac{1}{k^3}\right)=\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\frac{1}{3\Gamma(\omega)\Gamma(\omega^2)} \end{align} and hence the result
By the similar way we may get $\prod_{k=2}^{\infty}(1-\frac{1}{k^n})$