Convergence of series ${\sum_{n >= 1} \sqrt{n}\ln(\cosh(1/n))}$ The series is $${\sum_{n >= 1} \sqrt{n}\ln(\cosh(1/n))}$$
I´ve tried with the quotient criterion and with the limit criterion but the critiria does not decide. I know that this series converges but I can´t find an upper bound for the sequence.
 A: $$
\text{cosh}\left(\frac{1}{n}\right)\underset{(+\infty)}{=}1+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)
$$
Hence
$$
\ln\left(\text{cosh}\left(\frac{1}{n}\right)\right)\underset{(+\infty)}{=}\ln\left(1+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\underset{(+\infty)}{=}\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)
$$
You finally have the following asymptotic behaviour

$$
\sqrt{n}\ln\left(\text{cosh}\left(\frac{1}{n}\right)\right)\underset{(+\infty)}{\sim}\frac{1}{2n^{3/2}}$$

What do you know about $\displaystyle \sum_{n \geq 1}\frac{1}{n^{3/2}}$ ?
A: Observe that when $n\to +\infty,$
$$\ln (\cosh (\frac 1n))=\ln (1+\cosh (\frac 1n)-1) $$
$$\sim \cosh (\frac 1n)-1$$
$$= 2\sinh^2 (\frac {1}{2n})\sim \frac {2}{4n^2}\sim \frac {1}{2n^2}$$
thus
$$u_n=\sqrt {n}\ln (\cosh (\frac 1n))\sim \frac {1}{2n^{\frac 32}} $$
and $\sum u_n $ converges.
A: For any $x\geq 0$ we have $\log\cosh x\leq \frac{x^2}{2}$, hence your series is convergent and bounded by $\frac{1}{2}\zeta\left(\frac{3}{2}\right)$.
Actually
$$ \log\cosh\frac{1}{n} = \sum_{m\geq 0}\left(1+\frac{4}{\pi^2 n^2 (2m+1)^2}\right)=\sum_{m\geq 0}\sum_{h\geq 1}\frac{(-1)^{h+1} 4^h}{h \pi^{2h}n^{2h}(2m+1)^{2h}} $$
$$ \log\cosh\frac{1}{n} =\sum_{h\geq 1}\frac{(-1)^{h+1}(4^h-1)\zeta(2h)}{h\pi^{2h} n^{2h}}$$
$$ \sum_{n\geq 1}\sqrt{n}\log\cosh\frac{1}{n} =\sum_{h\geq 1}\frac{(-1)^{h+1}(4^h-1)\zeta(2h)\zeta(2h-1/2)}{h\pi^{2h}}$$
hence the original series can be bounded by $\frac{1}{2}\zeta\left(\frac{3}{2}\right)-\frac{1}{12}\zeta\left(\frac{7}{2}\right)+\frac{1}{45}\zeta\left(\frac{11}{2}\right)\leq\color{red}{\sqrt{5}-1}.$
