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

any hint?

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    $\begingroup$ 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. $\endgroup$ – Sasha Oct 1 '12 at 4:08
  • $\begingroup$ 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) $\endgroup$ – math st. Oct 1 '12 at 4:16
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Your hint is: It's the product of a lot of numbers each of which is between 0 and 1.

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

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