3
$\begingroup$

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

any hint?

  • 2
    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.2012-10-01
  • 0
    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)2012-10-01

1 Answers 1

2

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

  • 0
    So it is like $a_{n}<1$, then $a_{1}<1$, $a_{1}a_{2}2012-10-01
  • 0
    Can we find the exact value of the product?2012-10-01
  • 1
    try multiplying the first 5 terms by hand or with a calculator2012-10-01
  • 0
    very small number!2012-10-01
  • 0
    @MJD: But I think this method doesn't work if we have $\prod_{n=1}^{\infty}1+\frac{1}{n^{2}+1}$, right?2012-10-01
  • 0
    @mathst., I would have pointed out that all the numbers being multiplied are positive and no larger than $1/2.$2012-10-01