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

Question

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

Answer 1

$$\frac{1}{n^{\alpha}} \le \int_{n-1}^n \frac{1}{x^{\alpha}}dx $$ 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.

Answer 2

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.