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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