For which $a \in \mathbb{R}$ does the series $\sum_{k=1}^\infty \frac{1}{k^a}$ converge? I want to show for which $a \in \mathbb{R}$ the series $\sum_{k=1}^\infty \frac{1}{k^a}$ converges.
For $a = 0$ the series diverges
for $a < 0$ we have $\frac{1}{k^{-a}} = k^a$ and the series diverges as well, however I am not sure how to prove convergence/divergence for $a > 1$ ($a = 1$ is the harmonic series and also an upper bound for all $0 < a < 1$ so that should be sufficient)
So what I am asking is: how can I prove for which $ 1 < a < 2$ the series converges (since I know that $\sum_{k=1}^{\infty}\frac{1}{k^2}$ converges)
Any hints, ideas and feedback are welcome, thank you.
 A: One of the tests that can show that this series is convergent for $a>1$ is Cauchy condensation test:
For $(a_n)_{n\in\mathbb N}$ being a non-increasing sequence of non-negative numbers, the series $\sum_{n=1}^\infty a_n$ is convergent if and only if the series $\sum_{n=1}^\infty 2^n a_{2^n}$ is convergent.
For $a\ge 0$ the sequence $a_n = n^{-a}$ is non-negative and non-increasing , so we can apply this test. We have
$$ \sum_{n=1}^\infty 2^n a_{2^n} = \sum_{n=1}^\infty 2^n \frac{1}{2^{na}} = \sum_{n=1}^
\infty (2^{1-a})^n$$
This is a geometric series, it is convergent iff $2^{1-a}<1$, that is $a>1$.
A: For $a\leq 1$, you are correct, since for all $a\leq 1$, the series diverges. In fact, you don't need to split the three cases of $a=0$, $a<0$ and $0<a\leq 1$, because in all three cases, the harmonic series is a lower (you wrote upper, which was probably a typo) bound, and since the harmonic series diverges, so must all the other series.

For $a>1$, the easiest way of showing it would probably be the integral test for series convergence.
A: This is exactly same as "P-series test" theorem.
You can see the proof here
A: You may also use Cauchy Integration test.It is very easy to check that integration is possible only when $a>1.$
