First we transform the difference of two roots:

\(\displaystyle n^{\alpha}(\sqrt{n+1}-\sqrt{n})=n^{\alpha}(\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{n^{\alpha}}{\sqrt{n+1}+\sqrt{n}}. \)

For large values of \(n\) the denominator of the preceding fraction behaves roughly like \(\sqrt{n}\), and so we try to use the comparison test for

\(\displaystyle a_n=n^{\alpha}(\sqrt{n+1}-\sqrt{n})=\frac{n^{\alpha}}{\sqrt{n+1}+\sqrt{n}}\text{ a }b_n=\frac{n^{\alpha}}{\sqrt{n}}=n^{\alpha-\frac{1}{2}}. \)

Now \(\displaystyle \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{n^{\alpha}}{\sqrt{n+1}+\sqrt{n}}}{\frac{n^{\alpha}}{\sqrt{n}}}= \lim_{n\to\infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{2}\in(0,\infty). \)

So by the limit comparison test the given series converges iff the series \(\displaystyle \sum_{n=1}^{\infty}n^{\alpha-\frac{1}{2}}\) converges. So by the basic comparison test the given series converges iff \(\alpha-\frac{1}{2}<1\), which is the same as \(\alpha<\frac{1}{2}\).