I am taking honors Calculus II and have been doing reasonably well in the course until the current problem set which is due tomorrow. One exercise that is really giving me trouble is this:

Prove that $\sum\limits_{k=1}^\infty \frac{1}{k^5}$ is irrational.

It's easy to prove that the series converges by the p-test, but I I just don't see where to go with proving its irrationality. Any hints would be greatly appreciated, but please don't give the whole problem away.

I don't really know why my professor likes such problems, but we spent the last two lectures proving that things were irrational. We proved first that numbers like $\sqrt{2}$, $\log_2{3}$, and $\sqrt{2}+\sqrt{3}$ are irrational. Then last lecture we proved that $\pi$ is irrational.

But for the problem above, I tried to prove it by contradiction,

Suppose that $\sum\limits_{k=1}^\infty \frac{1}{k^5}$ is rational, then there exist $a,b$ such that $a \in \mathbb{Z}$ and $b \in \mathbb{Z}\setminus\{0\}$ such that $$\sum\limits_{k=1}^\infty \frac{1}{k^5} = \frac{a}{b}.$$ I really don't know where to go from here. I looked over my notes and we haven't done anything like this. The only thing that we did that was kind of similar was proving that $\sum\limits_{k=1}^n \frac{1}{k}$ is never an integer for $n > 1$, but this is only a finite series and it's dealing with an integer. Maybe there are related, but I don't know how I could use unique factorization to prove the above irrationality of the given series. Would that work?

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    $\begingroup$ Even the case of replacing the power $5$ by $3$ is not elementary to show irrational. See en.wikipedia.org/wiki/Ap%C3%A9ry's_constant $\endgroup$ – coffeemath Mar 13 '14 at 4:58
  • $\begingroup$ @coffeemath That's not particularly constructive. I'm taking honors calculus, so it would be pretty surprising that the last problem on our problem set would be "elementary". I don't mean to be hostile, but I am very tired as I have been working on this problem for ages. Do you have any suggestions on how to prove it? $\endgroup$ – Math is fun Mar 13 '14 at 5:03
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    $\begingroup$ My comment was meant in the spirit of: when the exponent $n=3$ the proof is extremely complicated, and when $n=5$ it would be even more so. See user134824's answer, for a backup on why at this point no one has a proof of irrationality for your problem. [also "elementary" doesn't mean easy, it means do-able by rather simple arguments using all calc, linear algebra, etc...] $\endgroup$ – coffeemath Mar 13 '14 at 5:07
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    $\begingroup$ @Mathisfun "Elementary" $\neq$ "Easy". See the "elementary" proof of the Prime Number Theorem. $\endgroup$ – Brian Fitzpatrick Mar 13 '14 at 5:07
  • $\begingroup$ Even the case of replacing the power of $5$ by the power of $2$ is very difficult, I suspect this question wasn't meant to be taken seriously. $\endgroup$ – JLA Mar 13 '14 at 5:35

This is an unsolved problem! See the Wikipedia article on Apéry's theorem. Zudilin showed in 2001 that at least one of the numbers $\zeta(5),\zeta(7),\zeta(9)$, and $\zeta(11)$ is irrational, and this appears to be the best result so far. (Here $\zeta(n)=\sum_{k=1}^\infty \frac{1}{k^n}$.)

  • $\begingroup$ I don't think this is true. I googled around and found this paper. I just need something a little simpler as that is way over my head. $\endgroup$ – Math is fun Mar 13 '14 at 5:10
  • $\begingroup$ That does seem to make me completely wrong, huh? I can't comment on the accuracy of that paper (it's over my head too), but in any case, the problem you were assigned is highly nontrivial. $\endgroup$ – user134824 Mar 13 '14 at 5:22
  • $\begingroup$ See en.wikipedia.org/wiki/Talk%3AAp%C3%A9ry%27s_theorem#Claim $\endgroup$ – Ted Mar 13 '14 at 5:29
  • $\begingroup$ I don't believe the arxiv paper can be trusted. I think it's safe to say that it's unknown whether or not $\zeta(5)$ is irrational. $\endgroup$ – nbubis Mar 13 '14 at 5:54
  • $\begingroup$ That's what I suspected. $\endgroup$ – user134824 Mar 13 '14 at 6:14

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