# Finding the infinite product $\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right)$

How do I find:

$$\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right) \quad$$

I am pretty sure that the infinite product converges, but if it doesn't please let me know if I have made an error. Also, could I have a nice explanation as to how would someone arrive to the answer.

Thanks alot.

-

An infinite product $\prod_{n \ge 0} (1 + u_n)$ where $u_n > 0$ converges if and only if $\sum_{n \ge 0} u_n$ converges, and $\sum_{n \ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}$ is a famous result by Euler.

Euler's product formula for $\sin z$ is hard to prove, but intuitive (it has roots at $\pm n \pi$): $$\frac{\sin z}{z} = \prod_{n \ge 1} \left( 1 - \frac{z^2}{\pi^2 n^2} \right)$$ This gives directly $\dfrac{\sin i}{i} = \dfrac{e - e^{-1}}{2}$ for your product

-
@PeterTamaroff, was fixing while you commented. Thanks! – vonbrand Apr 12 '13 at 17:30
OK. +1 ${}{}{}$ – Pedro Tamaroff Apr 12 '13 at 17:32
In short: $\sinh(1)$. – J. M. Apr 13 '13 at 16:48
@J.M., I found out too late :-( – vonbrand Apr 13 '13 at 16:54

The product converges because $\sum n^{-2}$ does. Can you see why? What is the greatest sum that can possibly appear when we expand the product?

In fact, suppose that $a_n\geq 0$ for each $n$. Set $$p_n=\prod_{k=1}^n a_k$$

Then $\log p_n=\sum_{k=1}^n\log a_k$

If $\sum a_k$ converges, then $a_k\to 0$, then since $$\lim_{x\to 0}\frac{\log (1+x)}x=1$$ so that $$\lim \frac{\log (1+a_n)}{a_n}=1$$

the comparison test tells us $\log p_n$ converges, say to $\ell$. By continuity of the logarithm, $p_n$ must converge to $p$ with $\log p=\ell$ so that $\lim p_n=e^\ell$.

Conversely, suppose $\log p_n=\sum_{k=1}^n\log a_k$ converges. This means that $\log(1+a_k)\to 0$ so that $a_k\to 0$. Comparison yields from $$\lim \frac{\log (1+a_n)}{a_n}=1$$ that $$\sum a_k$$ converges too. Thus, we have shown

PROP If $a_n\geq 0$ then $\prod (1+a_k)$ converges if and only if $\sum a_k$ (if and only if $\sum \log(1+a_k)$ does.)

-