Testing convergence of a series $\sum_{n=3}^\infty\ln\left(\frac{\cosh(\pi/n)}{\cos(\pi/n)}\right)$

Question

Given series $$\sum_{n=3}^\infty\ln\left(\frac{\cosh(\pi/n)}{\cos(\pi/n)}\right)$$ I should test convergence. I know, that I should use a comparison criterion. I tried expressing $$2\cosh(x)=e^x+e^{-x}$$ but got nowhere. What is the correct approach?

Answer 1

Recurring theme: you can use Taylor expansions, here of $\cos, \cosh$ and $x\mapsto\ln(1+x)$ when $x\to 0$. It's systematic, and it works.

In detail: $$\begin{align} \cosh \frac{\pi}{n} &= 1+\frac{\pi^2}{2n^2}+ o\left(\frac{1}{n^2}\right)\\ \cos \frac{\pi}{n} &= 1-\frac{\pi^2}{2n^2}+o\left(\frac{1}{n^2}\right)\\ \end{align} $$ so that $$ \begin{align} \ln \frac{\cosh \frac{\pi}{n}}{\cos \frac{\pi}{n}} &= \ln \frac{1+\frac{\pi^2}{2n^2}+ o\left(\frac{1}{n^2}\right)}{1-\frac{\pi^2}{2n^2}+ o\left(\frac{1}{n^2}\right)} = \ln \left(1+\frac{\pi^2}{2n^2}+ o\left(\frac{1}{n^2}\right) \right)\left(1+\frac{\pi^2}{2n^2}+ o\left(\frac{1}{n^2}\right)\right)\\ &=\ln \left(1+\frac{\pi^2}{n^2}+ o\left(\frac{1}{n^2}\right)\right)\\ &=\frac{\pi^2}{n^2}+ o\left(\frac{1}{n^2}\right)\\ \end{align}$$ so by comparison with the series $\sum_n \frac{1}{n^2}$ your series is convergent.

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We used, on top of the Taylor expansions of $\cos$ and $\cosh$ to second order, the low-order expansions of $\ln(1+u)$ and $\frac{1}{1+u}$ around $0$, all of them quite standard.