Convergence of $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$ I am checking for convergence of
$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$
First, we notice
$${n+1 \over n-1}= {1+{2 \over n-1}}$$
Then, we use $\ln(1+x) \le x$ and get
$$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1}$$
We now use $n-1 > {n \over 2}$
$${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1} < {1 \over \sqrt{n}}{2 \over {n \over 2}} = 4{1 \over n\sqrt{n}}$$
$\sum_{n=2}^{\infty}4{1 \over n\sqrt{n}}$ converges and so does $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$. Is this correct? Alternatively, is there a quicker way?

If we use the well-known equivalence,
$$\ln(1+X)\sim X\;\;(X\to 0),$$
we get that
$$\ln(1+\frac{2}{n-1})\sim\frac{2}{n-1}\sim\frac{2}{n}\;\;(n\to +\infty)$$
$$\implies \frac{1}{\sqrt{n}}\ln(\frac{n+1}{n-1})\sim\frac{2}{n^\frac{3}{2}}\;(n\to+\infty)$$
the general terms are positive and equivalent, the series are both convergent since $\frac{3}{2}>1$.
