Is this a valid solution to determine convergence / divergence to a series? I'm struggling a bit with determining convergence / divergence for series, and would just like to know if my solution below is valid solution. The reason I am asking is that this series is under the chapter with root test, ratio test and comparison test, but I solved it using just the criteria that if $\lim_{k\to\infty} ak = 0$ the series is convergent.
$\displaystyle\sum_{k=1}^{\infty} \left(\ln\frac{k+1}{k-1}-\frac{2}{k}\right)$
= $\displaystyle\lim_{k\to\infty} \left(\ln\frac{k\left(1+\frac{1}{k}\right)}{k(1-\frac{1}{k})}-\frac{2}{k}\right)$
= $\displaystyle\lim_{k\to\infty} \ln\left(1+\frac{1}{k}\right) - \lim_{k\to\infty} \ln\left(1-\frac{1}{k}\right) - \lim_{k\to\infty} \frac{2}{k} = 0$
Is this a valid solution?
 A: In order to have convergence of $\sum_{k\geq 1}a_k$ the fact that $\lim_{k\to +\infty}a_k=0$ is a necessary condition but not a sufficient one, as can be seen from the case $a_k=\frac{1}{k}$.
Additionally, $\ln\frac{k+1}{k-1}$ is not defined for $k=1$.
On the other hand
$$ S=\sum_{k\geq 2}\left(\ln\frac{k+1}{k-1}-\frac{2}{k}\right) = \sum_{k\geq 1}\left(\ln\left(1+\frac{2}{k}\right)-\frac{2}{k+1}\right)$$
is clearly convergent since $\ln\left(1+\frac{2}{k}\right)-\frac{2}{k}=O\left(\frac{1}{k^2}\right)$ and $\sum \frac{1}{k^2}$ is convergent.
In explicit terms
$$ S=2-2\gamma-\ln 2\approx 0.152421 $$
where $\gamma$ is the Euler-Mascheroni constant. You may also notice that
$$ \ln\frac{k+2}{2}-\frac{2}{k+1}=\int_{k}^{k+2}\frac{dt}{t}-\frac{2}{k+1}=\int_{-1}^{1}\left(\frac{1}{t+k+1}-\frac{1}{k+1}\right)\,dt $$
equals
$$ \int_{-1}^{1}\frac{-t}{(k+1)(t+k+1)}\,dt =2\int_{0}^{1}\frac{t^2}{(k+1)((k+1)^2-t^2)}\,dt$$
which is positive but less than $\frac{2}{3 k(k+1)(k+2)}$, such that $S\leq\frac{1}{6}$.
