Convergence of parameterized series $\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$ $$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$$
I guess that more useful form is:
$$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} + \sqrt{n})^{-p} \cdot \left( \ln(n-1) - \ln(n+1) \right) \right)$$
The question is, for what $ p \in \mathbb{R}$ does it converge?
I've made some experimental random checks, and there seems that there's something happening for -1 and -2, I think it converges for $p \in (-2; -1) \cup (-1; +\infty)$, but then again it's just some random test results.
How could I find for which $p$ it works, and which method of examining convergence should I use?
 A: HINT: Rewrite the series as
$$ \begin{eqnarray}
  \mathcal{S} &=& \sum_{n=2}^\infty \left( \sqrt{n+1}-\sqrt{n} \right)^p \cdot \ln \left(\frac{n-1}{n+1} \right) \\ &=& \sum_{n=2}^\infty \left( \frac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}} \right)^p \cdot \ln \left(1-\frac{2}{n+1} \right)\\  &=& \sum_{n=2}^\infty \frac{ \ln \left(1-\frac{2}{n+1} \right)}{\left(\sqrt{n+1} + \sqrt{n} \right)^p}
\end{eqnarray} $$
A: Writing it as
$$\sum\limits_{n = 2}^{ + \infty } {\left( {{{\left( {\sqrt {n + 1}  - \sqrt n } \right)}^p}\cdot\ln \left( {1 - {2 \over {n + 1}}} \right)} \right)} $$
and using $$\log \left( {1 - {2 \over {n + 1}}} \right) = {2 \over {n + 1}} + {2 \over {{{\left( {n + 1} \right)}^2}}} + o\left( {{1 \over {{n^3}}}} \right)$$
$$\sqrt {n + 1}  - \sqrt n  \sim {1 \over {2\sqrt n }}$$
means we're interested in how
$${1 \over {2{n^{p/2}}}}{2 \over {n + 1}} + {1 \over {2{n^{p/2}}}}{2 \over {{{\left( {n + 1} \right)}^2}}} + {1 \over {2{n^{p/2}}}}o\left( {{1 \over {{n^3}}}} \right)$$
behaves for large $n$. Can you take it from there?
A: Hint: 
$$1)\quad \frac{1}{ (\sqrt{n+1}+\sqrt{n})^m }\sim \frac{1}{ (\sqrt{n})^m }=\frac{1}{n^{m/2}}, $$
2) Use the integral test 
$$ \int_{2}^{\infty}\frac{\ln(x)}{x^a}dx $$ and find for what values of $a$ the integral converges.
