Convergence of $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ I'm trying to prove the convergence of $$ \sum_{n=1}^{\infty}\frac{1}{n^\alpha}$$
with $\alpha > 1$. 
For $\alpha \geq 2$ I can use the comparison test ($\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges) so I'm missing $2>\alpha>1$ and I'm pretty much out of ideas.
If you could offer some advice I would very much appreciate it.
Thanks in advance
 A: $$\frac{1}{n^{\alpha}} \le \int_{n-1}^n \frac{1}{x^{\alpha}}dx $$ 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.
A: 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.
A: 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.
