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$$ \sum \limits^{\infty }_{n=1}\left( \sum \limits^{n}_{k=0}k^{5}\right) ^{-1}$$ How can we put the series above in a nice closed form?

This problem was proposed on Brilliant.

Jack D'Aurizio
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user257567
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  • What is the context of the question? – Umberto P. Sep 17 '15 at 14:11
  • @UmbertoP. This above expression is a summation and inverse summation. My question how we put it in the closed form? – user257567 Sep 17 '15 at 14:14
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    I mean, where did the question come from? Sometimes knowing the context of a question helps in finding a solution. – Umberto P. Sep 17 '15 at 14:15
  • Closed form will be some real number (there are no free variables in it). Hard to find, I think. – drhab Sep 17 '15 at 14:18
  • Look at small powers first. $k^1$ has sum of $2$. $k^2$ has sum of $18-24 \log{2}$. And so on. – Ron Gordon Sep 17 '15 at 14:29
  • @GohP.iHan: please don't be disrespectful. I voted for reopening since the problem is really interesting, but you should moderate your language. – Jack D'Aurizio Sep 18 '15 at 09:56
  • @jack: Deleted comment. I was furious because he copied one of the incomplete solutions (which was deleted already) from here, word for word. If he haven't done so. I wouldn't have made any comments. – GohP.iHan Sep 18 '15 at 11:14
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    @GohP.iHan: I understand your disappointment. Anyway, the flag system gives a good way for dealing with such issues. – Jack D'Aurizio Sep 18 '15 at 11:21
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    @JackD'Aurizio: Thanks for your reply. And I would like to thank you for indirectly answering my question. I did the tedious partial fraction and logarithmic Weierstrass product derivative approach (without Digamma). Your solution is an eye opener. – GohP.iHan Sep 18 '15 at 11:24

3 Answers3

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So, we want to compute:

$$ \sum_{n\geq 1}\frac{12}{n^2(n+1)^2(2n^2+2n-1)}.\tag{1}$$ By using the digamma reflection formula we have: $$ \sum_{n\geq 1}\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)\frac{1}{2n^2+2n-1}=7-\frac{\pi^2}{3}+\frac{4\pi}{\sqrt{3}}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right) \tag{2}$$ hence the problem boils down to computing: $$ \sum_{n\geq 1}\frac{1}{n(n+1)(2n^2+2n-1)}=1+\frac{\pi}{\sqrt{3}}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right)\tag{3} $$ with the same technique, i.e. through the identities: $$ \sum_{n\geq 1}\left(\frac{1}{n+a}-\frac{1}{n+b}\right)= \psi(a+1)-\psi(b+1),\qquad \sum_{n\geq 1}\frac{1}{(n+a)^2}=\psi'(a+1)$$ and the well-known $\psi(z)-\psi(1-z)=-\pi\cot(\pi z)$. By putting all together, we reach:

$$ \sum_{n\geq 1}\left(\sum_{k=1}^{n}k^5\right)^{-1}=60-4\pi^2+8\pi\sqrt{3}\,\tan\left(\frac{\pi\sqrt{3}}{2}\right).\tag{4}$$

We may compute the non-trivial series: $$\sum_{n\geq 1}\frac{1}{(2n+1)^2-3}$$ also by considering the logarithmic derivative of the Weierstrass product for the cosine function. That gives: $$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-a^2}=\frac{\pi}{4a}\,\tan\left(\frac{\pi a}{2}\right).\tag{5}$$

Jack D'Aurizio
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  • $$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-a^2}=\frac{\pi}{4a},\tan\left(\frac{\pi a}{2}\right).\tag{5}$$. What is the name ofThis summation sir? – user257567 Sep 17 '15 at 17:08
  • Thanks, Jack. This question was actually made by me which this OP ripped off. The only difference between your solution and mine was I didn't use the digamma function property. – GohP.iHan Sep 17 '15 at 18:05
  • Jack: Do you think that it's possible to sum for powers of 4 instead of powers of 5? I tried it but without success. – GohP.iHan Sep 17 '15 at 18:06
  • @user257567: I do not know if there is a specific name for such identity, anyway it is well-known. It can be proved along the lines I gave above (i.e. through the Weierstrass product for the cosine function) or through some manipulation of Fourier series. – Jack D'Aurizio Sep 18 '15 at 09:50
  • @GohP.iHan: if we replace $k^5$ with $k^4$ the answer depends on $\psi\left(\frac{3}{2}\right)$ and on $\psi\left(\frac{9\pm\sqrt{21}}{6}\right)$. In such a case, however, it is not possible to write the sum in a nice form like $(4)$. – Jack D'Aurizio Sep 18 '15 at 09:52
  • Thanks. I'm not well-verse in digamma function properties. Glad I got the confirmation. – GohP.iHan Sep 18 '15 at 11:10
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  1. Write the inside summation in a closed form.
  2. Invert this expression.
  3. Use partial fraction decomposition.
  4. Evaluate the infinite summation. There should be some telescoping.
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Hint:

let S be the closed form of our summation .

$$\sum_{k=1}^nk^5=\frac{1}{6}\sum_{j=0}^{5}\binom{6}{j}B_jn^{6-j}=\frac{1}{36}(6n^6-18n^5+15n^4-3n^2)=\frac{1}{12}(n^2(1+n)^2(2n(1+n)-1))$$

where is $B_j$ is Bernoulli number.

so

$$S=12\sum_{n=1}^{\infty}\frac{1}{n^2(1+n)^2(2n(1+n)-1)}$$

$$S=12\sum_{n=1}^{\infty}\left (\frac{-1}{n^2}-\frac{1}{(n+1)^2}+\frac{4}{(2n(1+n)-1)} \right )$$

$$S=4(15-\pi^2+2\sqrt{3}\tan(\frac{\sqrt{3}\pi}{2}))$$

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mnsh
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  • Why the last line? Was it copied from Mathematica or a similar program, with no relation to the computations before it? – Did Sep 02 '16 at 08:07
  • @Did ,Sorry cuz i didnt give more explanation ,u know that for Zeta(2) ,and for the 2nd sum i can get it using pole expansion of $\tan \frac{\pi}{2}z$ – mnsh Sep 02 '16 at 17:29
  • so it related to the sum before it ,i said that is a hint and if any one ask me to explain more i will. – mnsh Sep 02 '16 at 17:31