0
$\begingroup$

Let $a$ be a complex constant with norm $|a| > 1$. Show that the series $$\sum_{k=1}^\infty \frac{a^k}{k^4}$$ diverges. The problem is really easy if $a \in \mathbb R$ is real: one just have to notice that the term does not tend to zero and therefore the necessary condition for series convergence isn't fulfilled. But what if $a$ is allowed to be a complex value?

[Added] My teacher said that the ratio test is not enough and we shall show more, that

$$\lim_{n \to \infty} \left| \sum_{k=1}^n \frac{a^k}{k^4}\right| = + \infty.$$

Could someone explain to me what he really meant (e.g. why the ratio test was insufficient in this case)?

$\endgroup$
  • 1
    $\begingroup$ "one just have to notice that the term does not tend to zero" ...And the argument works as well when $a$ is complex with $|a|\gt1$, no? $\endgroup$ – Did Mar 2 '15 at 17:22
  • 1
    $\begingroup$ This is a polylogarithm. $\endgroup$ – Lucian Mar 2 '15 at 20:14
3
$\begingroup$

If we let $b_k = \frac{a^k}{k^4}$ then we have by the ratio test $$\begin{align} \lim_{k \to \infty }\left|\frac{b_{k+1}}{b_k}\right| =\lim_{k \to \infty }\left|\frac{a^{k+1}}{(k+1)^4}\cdot \frac{k^4}{a^k}\right| \\ = |a|\left( \lim_{k \to \infty }\left|\frac{k^4}{(k+1)^4}\right|\right) \\ = |a|>1 \end{align}$$ so it doesn't matter how large or small the real or complex part of $a$ is, since we assumed $|a|>1$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That's how I was trying to prove the divergence. Now I have added a clarification to my post, could you take a look? $\endgroup$ – user207868 Mar 2 '15 at 18:03
  • $\begingroup$ @LeonAragones I am not seeing yet why the ratio test is not enough. Why did your teacher say this? I double-checked the wikipedia page about it, and the ratio test should be conclusive for a sum of complex values like this one. How about the root test in the other answer of your question? $\endgroup$ – graydad Mar 2 '15 at 18:05
0
$\begingroup$

Since

$$\lim_{k\to \infty} \left|\frac{a^k}{k^4}\right|^{1/k} = \lim_{k\to \infty} \frac{|a|}{k^{4/k}} = |a| > 1,$$

your series diverges by the root test.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Diverges by the root test, I believe you mean. $\endgroup$ – Joel Mar 2 '15 at 17:22
  • $\begingroup$ @Joel I'm just tired. Need to get rest. :) $\endgroup$ – kobe Mar 2 '15 at 17:22
  • $\begingroup$ I have had those days. Cheers :) $\endgroup$ – Joel Mar 2 '15 at 17:23

Your Answer

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