Testing convergence of series $\sum_{n=2}^\infty\frac{\ln\left(\frac{n+1}{n-1}\right)}{\sqrt{n}}$ Lets have a series
$$\sum_{n=2}^\infty\frac{\ln\left(\frac{n+1}{n-1}\right)}{\sqrt{n}}$$
However, I have absolutely no clue how to try to continue. I could probably use the integral criterion and integrate the problem using the residue theorem, but that is too much of a hassle. Is there an easy way to prove the convergence of this series?
 A: Notice that
$$\ln\left(\frac{n+1}{n-1}\right)=\ln\left(1+\frac2{n-1}\right)\sim\frac2{n-1}\sim\frac2n$$
so the given series is convergent by asymptotic comparaison with the series $\sum\frac1{n\sqrt n}$.
A: If you are familiar with Taylor expansions, and the limit comparison test:
The general term of your series is, when $n\to \infty$,
$$
\frac{\ln\left(\frac{n+1}{n-1}\right)}{\sqrt{n}}
= \frac{\ln\left(1+\frac{2}{n-1}\right)}{\sqrt{n}}
= \frac{\frac{2}{n}+ o(\frac{1}{n})}{\sqrt{n}}
= \frac{2}{n^{3/2}}+ o\left(\frac{1}{n^{3/2}}\right)
$$
so by the limit comparison test you series converges (at the same rate as the series $\sum_n \frac{2}{n^{3/2}}$.)

We used the facts (low-order Taylor expansions) that, when $u\to 0$,
$$
\ln(1+u) = u+o(u), \qquad\text{and} \qquad \frac{1}{\frac{1}{u}-1} = u + o\left(u\right)
$$
A: For example, (limit or not) comparison:
$$\frac{n+1}{n-1}=1+\frac2{n-1}\implies\log\left(1+\frac2{n-1}\right)\le\frac2{n-1}\implies$$
$$\frac{\log\frac{n+1}{n-1}}{\sqrt n}\le\frac2{(n-1)\sqrt n}$$
and the series converges by (limit, say) comparison with $\;\sum\frac1{(n-1)\sqrt n}\;$
