Convergence of $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$

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.

$$\frac{1}{n^{\alpha}} \le \int_{n-1}^n \frac{1}{x^{\alpha}}dx $$ for $\alpha \gt 1$ so
$$\sum_{n=1}^{\infty}\frac{1}{n^\alpha} = 1 + \sum_{n=2}^{\infty}\frac{1}{n^\alpha} \le 1+\int_{1}^\infty \frac{1}{x^{\alpha}}dx = 1+ \frac{1}{\alpha-1}.$$

Since each term is positive and the sum is bounded above, the series is convergent.

As the sequence $\displaystyle<n^\alpha>_{n=1}^{\infty}$ is decresing we can use the integral test to check its convergence.

$\displaystyle\int_{1}^{\infty}\frac{1}{x^\alpha}dx=1$ for all $\alpha>1$.

Hence the series is convergent.