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

Compute:

\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??

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    @jspecter but this series ain't same as the one for $\zeta(2)$.2012-01-16

2 Answers 2

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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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Actually there is a neat form for the partial products.

Let

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

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

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    This is what we would call a "telescoping" infinite product.2012-01-16