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Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$


\begin{align*} \lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2}) \end{align*}

Well, I do so: $$\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\lim_{n\to+\infty}\prod_{j=2}^n (1-\frac{1}{j^2})$$ let $$a_n = \prod_{j=2}^n (1-\frac{1}{j^2})\quad\Rightarrow\quad \ln a_n = \ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)$$ so: $$\ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)=\sum_{j=2}^{\infty} \ln (1-\frac{1}{j^2})$$ consider $$\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^2})$$ but as you study this series??


marked as duplicate by Martin Sleziak, Listing, Rudy the Reindeer, David Mitra, jspecter Jan 16 '12 at 13:42

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  • $\begingroup$ @jspecter but this series ain't same as the one for $\zeta(2)$. $\endgroup$ – nb1 Jan 16 '12 at 12:13

Actually there is a neat form for the partial products.


$$f(n)=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\prod _{k=2}^n \left(1-\frac{1}{k^2}\right)$$

We show by induction that $f(n)=\frac{n+1}{2 n}$. $$f(n)=\left(1-\frac{1}{n^2}\right)f(n-1)=\frac{n^2-1}{n^2}\cdot\frac{n}{2(n-1)}=\frac{n+1}{2n}$$

Therefore $\lim_{n\to+\infty}f(n)=\lim_{n\to+\infty}\frac{n+1}{2 n}=\lim_{n\to+\infty}(\frac{1}{2}+\frac{1}{2n})=\frac{1}{2}$

  • $\begingroup$ I think the last "$=f(n+1)$" is wrong. $\endgroup$ – draks ... Jan 16 '12 at 12:31
  • $\begingroup$ Thank you for the correction. I fixed it $\endgroup$ – Listing Jan 16 '12 at 12:33
  • $\begingroup$ This is what we would call a "telescoping" infinite product. $\endgroup$ – GEdgar Jan 16 '12 at 16:16

If you write it like $$ \lim_{n\to \infty}\prod_{j=2}^n \frac{(j+1)(j-1)}{j^2} $$ you get $$ \lim_{n\to \infty}\frac{1}{2}\frac{(n+1)!(n-1)!}{(n!)^2}, $$ where factorials cancel in the limit and give $\frac{1}{2}$, because a $2$ is missing in the numerator to get $(n+1)!$.


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