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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?

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See formula 20 and formula 27 here. –  J. M. Jul 9 '12 at 18:35
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@Chris'sSister - off topic - how do you come up with these nice problems? In short would you recommend THE books on calculus with questions like these and how to solve them. Practically unique emphasis on limits, integration and series. Thanks –  nt.bas Dec 26 '12 at 13:28
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@nt.bas Hi. I'm very glad to read that you enjoy my questions. Some of the questions simply come from my mind (major part of them), others are just modified versions of some problems that I find interesting, and a few of them are problems met in various publications related to math contests. Math SE is one of the greatest virtual books I'd recommend because you may find here very nice calculus problems with very nice solutions. I hope that helps! :-) Chris. –  Chris's sis Dec 26 '12 at 13:52
    
@Chris'ssister : Thanks that is helpful! Actually I was trying your questions but I couldn't solve many of them. Therefore I would very much like to know which calculus books would help build the kind of intuition you have in addition to math SE, of course. My goal is to read a chapter in the book(s) then come back here (on math SE) and pick problems to try. Most books I've read (Spivak included) won't mention gamma and beta functions, limits of sums(products), the zeta function, asymptotic analysis,... which are used frequently in solutions. Thanks again! –  nt.bas Dec 26 '12 at 14:08
    
@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! –  nt.bas Dec 26 '12 at 14:38
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2 Answers 2

up vote 14 down vote accepted

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.

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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 $\prod_{k=2}^{\infty}(1-\frac{1}{k^n})$

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