Convergence of $\sum_{n=2}^{\infty}(-1)^nn\ln^2\left(\frac{n+1}{n-1}\right)$ I'm looking for help checking if the series:
$$\sum_{n=2}^{\infty}(-1)^nn\ln^2\left(\frac{n+1}{n-1}\right)$$
converges absolutely, conditionally or diverges.  
Here are my thoughts:
For conditional convergence I should use Leibnitz theorem.
I can show that $\lim_\limits{{n\to \infty}} a_n =0$ but when I tried to show it is monotonous I hit a dead end (I tried by finding the derivative or looking at $a_n- a_{n+1}$).
For absolute convergence I tried to use different tests but got limit of 1 so I couldn't learn from that.
Any ideas?  
 A: $$\log\frac{n+1}{n-1} = \log\frac{1+\frac{1}{n}}{1-\frac{1}{n}} = 2\,\text{arctanh}\frac{1}{n} = \frac{2}{n}+O\left(\frac{1}{n^3}\right) \tag{1}$$
leads to
$$ n\log^2\frac{n+1}{n-1} = \frac{4}{n}+O\left(\frac{1}{n^3}\right) \tag{2} $$
so the given series is conditionally convergent but not absolutely convergent, i.e. has exactly the same convergence behaviour of $4\sum_{n\geq 2}\frac{(-1)^n}{n}$, since $\sum_{n\geq 2}\frac{1}{n^3}$ is clearly absolutely convergent.
Are you interested in a exact evaluation, too?
A: $\ln \left(\frac{n+1}{n-1}\right) = \frac{2}{n-1} + O\left(\frac{1}{n^2}\right)$. Can you conclude from that ? 
A: $$\ln (n+1)=\ln (n)+\ln (1+1/n) $$
$$=\ln (n)+\frac {1}{n}-\frac {1}{2n^2}+\frac {1}{3n^3}(1+\epsilon (n)) $$
$$\ln (n-1)=\ln (n)-\frac {1}{n}-\frac {1}{2n^2}-\frac {1}{3n^3}(1+\epsilon (n)) $$
thus
$$\ln (\frac {n+1}{n-1})=\frac {2}{n}+\frac {2}{n^3}(1+\epsilon (n)) $$
and
$$\ln^2 (\frac {n+1}{n-1})=\frac {4}{n^2}+\frac {8}{n^4}(1+\epsilon (n)) $$
hence
$$u_n=\frac {(-1)^n4}{n}+\frac {(-1)^n8}{n^3}(1+\epsilon (n)) $$
the sum of a conditionally convergent $( \sum(-1)^n\frac {4}{n} ) \;$ and an absolutely convergent $( \sim \frac {8}{n^3}) \;$  is a conditionally convergent series.
