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The following question is from my analysis assignment and I was unable to solve it.

Find the value of $\prod_{n=1}^{\infty}(1+ 1/n^2)$ .

I tried to do some algebraic manipulation but I was not sucessful in that.Also, If the terms doesn't appear to me to cancel in a fashion.

I am at loss of ideas so I thought I should post here.

1 Answers1

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$$\prod_{k=1}^{\infty} \left(1+\frac{z^2}{k^2}\right)=\frac{\sinh \pi z}{\pi z}.$$ Put $z=1$ to get $$\prod_{k=1}^{\infty} \left(1+\frac{1} {k^2}\right)=\frac{\sinh \pi}{\pi}$$

Edit: An Elementry Proof:

$$F_n(x)= x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)......(x-n\pi)(x+n\pi)$$ $\sin x$, is finite and has infinitely many zero. $F_n(x)$ has $2n+1$ zeros but as $n\to \infty$ $\lim_{n\to \infty} F_n(x)\to \infty$, so $F_n$ cannot represent $\sin x$. But $$ G_n(x)=x(1-\frac{x}{\pi})(1+\frac{x}{\pi}) (1-\frac{x}{2\pi})(1+\frac{x}{2\pi}).....(1-\frac{x}{n\pi})(1+\frac{x}{n\pi}).~~~~~(1)$$ So $G_n$ can represent $\sin x$ when $G_n$ converges as $n\to \infty$, it also gives $\frac{\sin x}{x}=1$ when $x \to 0$. Thus Eq. (1) gives the correct representation of $$\frac{\sin x}{x}=\prod_{n=1}^{\infty} \left(1-\frac{x^2}{n^2\pi^2}\right) \implies \frac{\sinh \pi z}{ \pi z}=\prod_{n=1}^{\infty} \left(1+\frac{z^2}{n^2}\right) $$ Here we took $x=iz.$

Z Ahmed
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  • I amalso asked to find the infinite product of $\prod_{n=1}^{\infty} (1+1/n^2 + 1/n^4)$ . Does this construction given by you in answer will help or that's entirely different approach?, Thought of commenting In case you know it. –  Dec 15 '20 at 13:15