Convergence of infinite product and its limit

Question

I wanted to find $\prod_{n=2}^{\infty}(1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+...)$ and ended up simplifying it as $\prod_{n=2}^{\infty} \frac{n^2}{n^2-1}$. Now the partial product is $\frac{2n}{n+1}$ it converges and it's limit is 2.. am I correct? Kindly share your views

Answer 1

We can check your results, in order to verify your answer.

• First, we can show that $$1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots =\sum_{k=0}^{+\infty}\frac{1}{n^{2k}}.$$
• If $n\geqslant 1$, we can show that $$\sum_{k=0}^{+\infty}\frac{1}{n^{2k}}=\sum_{k=0}^{+\infty}\left(\frac{1}{n^2}\right)^k=\frac{n^2}{n^2-1}$$
• Then, $$\prod_{n=2}^{+\infty}\left(1+\frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots \right)=\prod_{n=2 }^{+\infty}\sum_{k=0}^{+\infty}\left(\frac{1}{n^2}\right)^k=\prod_{n=2}^{+\infty}\left(\frac{n^2}{n^2-1}\right)$$
• If you showed that for all $N\in\mathbf{N}$ we can write $$\prod_{n=2}^{N}\left(\frac{n^2}{n^2-1}\right)=\frac{2N}{N+1},$$ (this is indeed true, but you should have to justify it; a way it is using partial fraction), then taking $N\to +\infty$ both sides we find $$\prod_{n=2}^{+\infty}\left(\frac{n^2}{n^2-1}\right)=\lim_{N\to +\infty}\prod_{n=2}^{N}\left(\frac{n^2}{n^2-1}\right)=\lim_{N\to +\infty}\left(\frac{2N}{N+1}\right)=2$$ (it follows directly by definition) and then we get the answer $2$ as you said.