<s>[INST] 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.
 [/INST] 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 $\alpha > 0$.
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