# 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.

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 $$a>1$$.

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, 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.

• That sounds like a good idea however the class I am taking has not yet talked about integrals yet so I would have to proof the integral test to be able to use it, is there a way of doing that another way? – Pierre May 15 at 7:57
• @Pierre Integrals are pretty basic tools when it comes to analysis. I can't currenlty think of a simple way of proving what you need, sorry. – 5xum May 15 at 8:01
• Without using integrals, the Cauchy condensation test also works for $a>1$. – Adam Latosiński May 15 at 8:48

This is exactly same as "P-series test" theorem.

You can see the proof here

You may also use Cauchy Integration test.It is very easy to check that integration is possible only when $$a>1.$$