<s>[INST] 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?
 [/INST] You are right. Alternatively you can use the asymptotic comparison test by noting that
$${1 \over \sqrt{n}}\ln\left({n+1 \over n-1}\right)=
{1 \over \sqrt{n}}\ln\left({1+{2 \over n-1}} \right)\sim \frac{2}{n\sqrt{n}}=\frac{2}{n^{3/2}}$$
where we used that $\ln(1+x)\sim x$ when $x\to 0^+$.
It is similar to your approach but you do not deal with inequalities.
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