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Would you please give hint for these two(not homework)~~ $$\Pi_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$$ and $$\Pi_{1}^{\infty}(1+\frac{}{})e^{\frac{-z}{n}} \text{ converges absolutely and uniformly on every compact set }$$

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    Take the logarithm of the series.2012-04-19
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    Suggestion: look at the finite products (from 1 to, say, $m$) and apply log. Then let $m\rightarrow \infty$.2012-04-19
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    Related to first one: http://www.ams.org/bookstore/pspdf/gsm-97-prev.pdf2012-04-19

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For $(1)$, note that $$1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{n-1}{n}\frac{n+1}{n}.$$

Now write down the first few terms of the product, in factored form. (So each term of the original product is represented as the product of two terms.) We get $$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\cdots.$$ Note that second and third term cancel, as do fourth and fifth, and so on. Almost everything dies, and we are left with $\frac{1}{2}$.

This does not quite complete things, we should be more formal about the limiting process. But we can use the cancellation to produce an explicit formula for $$\prod_{k=2}^n \left(1-\frac{1}{k^2}\right).$$ Then it is easy to show that for large $n$ the above product is not far from $\frac{1}{2}$.