5
$\begingroup$

Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.

$\endgroup$

closed as off-topic by user147263, Martin R, Did, quid, J. W. Perry May 17 '15 at 4:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Martin R, Did, quid, J. W. Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Use the formula for Geometric series first. Then multiply. $\endgroup$ – Mann May 16 '15 at 6:34
  • 2
    $\begingroup$ Each one of those series is geometric, so you can replace it with a nice closed form. $\endgroup$ – Gregory Grant May 16 '15 at 6:34
  • $\begingroup$ oh i missed that one $\endgroup$ – Nebo Alex May 16 '15 at 6:49
  • $\begingroup$ After noticing that brackets contain geometric series and simplifying you get to a problem which has been asked on this site before. See, for example, this question and some of the questions linked to it. $\endgroup$ – Martin Sleziak May 16 '15 at 9:06
13
$\begingroup$

The product is

$$\prod_{n=2}^{\infty} \frac1{1-\frac1{n^2}} = \prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \prod_{n=2}^{\infty} \frac{n^2}{(n-1)(n+1)} = \frac{2 \cdot 2}{1 \cdot 3}\cdot \frac{3 \cdot 3}{2 \cdot 4}\cdot \frac{4 \cdot 4}{3 \cdot 5} \cdots = 2$$

$\endgroup$
7
$\begingroup$

Step $1$: Notice that $$\prod_{n=2}^{\infty} \bigg(\sum_{k=0}^{\infty} \frac{1}{n^{2k}}\bigg) = \prod_{n=2}^{\infty} \frac{1}{1-\frac{1}{n^2}} = \prod_{n=2}^{\infty}\frac{n^2}{n^2-1}$$

Step $2$: Use induction on $N$ to show that $$\prod_{n=2}^N \frac{n^2}{n^2-1} = \frac{2N}{N+1}$$

Step 3: Conclude that $$\prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \lim_{N \to \infty} \frac{2N}{N+1} = 2$$

$\endgroup$
5
$\begingroup$

The product is $$P=\prod_{n\ge 2}\frac{1}{1-1/n^2}=\lim_{x\to 1}\frac{\pi x(1-x^2)}{\sin \pi x}=2$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.