# how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? [closed]

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.

• Use the formula for Geometric series first. Then multiply. Commented May 16, 2015 at 6:34
• Each one of those series is geometric, so you can replace it with a nice closed form. Commented May 16, 2015 at 6:34
• oh i missed that one Commented May 16, 2015 at 6:49
• 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. Commented May 16, 2015 at 9:06

The product is

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

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 $$\prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \lim_{N \to \infty} \frac{2N}{N+1} = 2$$

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