Sorry to dig this up, but I would like to add that there is another way to do it without integral comparison test.
Write $\alpha=1+\beta$. Then $\beta>0$. Then group the summands in terms of powers of two and estimate by a geometric series:
$\sum_{n=3}^\infty \frac{1}{n^\alpha}=\underbrace{\frac{1}{3^\alpha}+\frac{1}{4^\alpha}}_{\le \frac{1}{2^\alpha}+\frac{1}{2^\alpha}=2\frac{1}{2^\alpha}=\frac{1}{2^\beta}}+\underbrace{\frac{1}{5^\alpha}+\frac{1}{6^\alpha}+\frac{1}{7^\alpha}+\frac{1}{8^\alpha}}_{\le 4\cdot \frac{1}{4^{\alpha}}=\frac{1}{4^\beta}}+\cdots\le \sum_{k=1}^\infty \frac{1}{2^{k\beta}}=\frac{2^{-\beta}}{1-2^{-\beta}}<\infty$
Edit: This is the Cauchy condensation test as pointed out below.