# Infinite product convergence

Reading some notes I have taken during class, I am unable to see the connection between these lines: $g(z)=\prod_{k\geqslant 0}(2z\lambda_k + 1)$, where $\lambda_k=4/((2k+1)^2 \pi^2)$

And

The zeroes of $h(z)=g(z^2)$ are $(i/\sqrt{2\lambda_k})$ (ok, why not) which implies $g(z^2)=\sinh(\sqrt{2z})$ (how?).

Appreciate any hints.

-
Could you define the numbers $\lambda_k$ for us? – Shaun Ault Dec 9 '12 at 20:14
yes, you already pointed me to something ;-) $\lambda_k=4/((2k+1)^2\pi^2)$ – citrucel Dec 9 '12 at 20:18
Here is the full text the zeroes of $h(z)=g(z^2)$, viewed as a function of the complex variable z, are $z_k=i/\sqrt(2\lambda_k)=1/4\sqrt2(2k+1)\pi$ this implies that $g(z^2)=sinh(\sqrt2z)$ – citrucel Dec 9 '12 at 20:31