Find the nature of $\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha \ln \frac{n + 1}{n - 1}$ I need to find whether the following series converges or diverges:
$$\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha \ln \frac{n + 1}{n - 1}$$
By graphing the sum, it seems that it converges if and only if $\alpha \gt 1$. I thought that Dirichlet's Criterion is the most suitable for this series, because $\ln \frac{n + 1}{n - 1}$ is decreasing and $\lim_{n \rightarrow \infty} \ln \frac{n + 1}{n - 1} = 0$. But $\sum_{n = 2}^\infty (\sqrt{n + 1} - \sqrt{n})^\alpha$ seems to be divergent regardless of $\alpha$, so I can't use this idea.
 A: Hint: $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$, implying
$
 (n+1)^{-1/2} <2(\sqrt{n+1}-\sqrt{n} )<  n^{-1/2}
$. Further, $\log((n+1)/(n-1)) = \log(1+2/(n-1)$; using Bernoulli's inequality we have $e^x\approx  1+x$ for small $x$, so we can compare to $2/(n-1)\approx 2/n$. Can you take it from here?
A: You have $$(\sqrt{n + 1} - \sqrt{n})^\alpha \ln \left(\frac{n + 1}{n - 1} \right)= (\sqrt{n})^{\alpha}\left(\sqrt{1 + \frac{1}{n}} - 1\right)^\alpha \ln \left(1 +\frac{2}{n - 1} \right)$$
$$ \sim (\sqrt{n})^{\alpha}\left(\frac{1}{2n}\right)^\alpha\frac{2}{n-1} = \frac{2}{(2\sqrt{n})^{\alpha}(n+1)} \sim \frac{2^{1-\alpha}}{n^{1+\frac{\alpha}{2}}}$$
So the series converges iff $1+\frac{\alpha}{2} > 1$, i.e. iff $\alpha > 0$.
A: We have that
$$\left(\sqrt{n + 1} - \sqrt{n}\right)^\alpha =\left(\frac1{\sqrt{n + 1} + \sqrt{n}}\right)^\alpha\sim \frac1{n^\frac \alpha 2} $$
and
$$n\ln \left(\frac{n + 1}{n - 1}\right)=n\ln \left(1+\frac{2}{n - 1}\right) \to 2$$
therefore
$$(\sqrt{n + 1} - \sqrt{n})^\alpha \ln \left(\frac{n + 1}{n - 1}\right) \sim \frac1{n^{1+\frac \alpha 2}}$$
and we can determine the convergence by limit comparison test.
