2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $a>1$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Pierre May 15 at 7:57
  • $\begingroup$ @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. $\endgroup$ – 5xum May 15 at 8:01
  • 1
    $\begingroup$ Without using integrals, the Cauchy condensation test also works for $a>1$. $\endgroup$ – Adam Latosiński May 15 at 8:48
1
$\begingroup$

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

You can see the proof here

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.