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Does this product converge? $$\prod_{n=1}^{\infty}\frac{1}{n^{2}+1}$$

any hint?

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It is depends on your definition of convergence of the product. It is not uncommon to say that this product diverges to zero. This is because the sum $\sum_{n=1}^\infty \log\left(\frac{1}{1+n^2}\right)$ is divergent. – Sasha Oct 1 '12 at 4:08
I was confused about this fact: The product of positive real numbers$ a_n<1 $, $ \prod_{n=1}^{\infty} a_n$ converges if and only if the sum $\sum_{n=1}^{\infty} \log a_n$ converges. So how this doesn't contradicts MJD answer below? (btw, by converge I mean finite) – math st. Oct 1 '12 at 4:16

Your hint is: It's the product of a lot of numbers each of which is between 0 and 1.

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So it is like $a_{n}<1$, then $a_{1}<1$, $a_{1}a_{2}<a_{2}<1$, $a_{1}a_{2}a_{3}<a_{3}<1$, and so on! – math st. Oct 1 '12 at 4:10
Can we find the exact value of the product? – math st. Oct 1 '12 at 4:11
try multiplying the first 5 terms by hand or with a calculator – Jonathan Oct 1 '12 at 4:22
very small number! – math st. Oct 1 '12 at 4:29
@MJD: But I think this method doesn't work if we have $\prod_{n=1}^{\infty}1+\frac{1}{n^{2}+1}$, right? – math st. Oct 1 '12 at 4:42

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