For what a will this series converge? $\sum_{n=1}^{\infty}n^a\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$
I tried to use the D'Alembert ratio test and it didn't quite work out. The series inside is telescoping but I don't know if that information will be useful. Can someone help?

We have
$$
\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}} =  \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} + \sqrt{n})} \sim \frac{1}{2n^{3/2}}.
$$
Hence $a_n = n^{\alpha} \Bigl( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \Bigr) \sim \frac{1}{2n^{3/2-a}}$. The sum $\sum_n a_n$ is convergent iff $\sum_n \frac{1}{n^{3/2-a}}$ is convergent. Using integral test for convergence we get that $\sum_n \frac{1}{n^{3/2-a}}$ is convergent iff $3/2-a > 1$.
So, the sum $\sum_n a_n$ is convergent iff $a<\frac12$.
