I was recently told to compute some integral, and the result turned out to be a scalar multiple of the series $$\sum\limits_{n=1}^\infty\frac{1}{(n(n+1))^p},$$ where $p\geq 1$. I know it converges by comparison for $$\dfrac{1}{(n(n+1))^p}\leq\dfrac{1}{n(n+1)}<\dfrac{1}{n^2},$$ and we know thanks to Euler that $$\sum\limits_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.$$ I managed to work out the cases where $p=1$ and $p=2$. With $p=1$ being a telescoping sum, and my solution for $p=2$ being $$\frac13\pi^2-3,$$ which I obtained based on Euler's solution to the Basel Problem. I see no way to generalize the results to values to arbitrary values of $p$ however. Any advice on where to start would be much appreciated.

Also, in absence of another formula, is the series itself a valid answer? Given that it converges of course.

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    $\begingroup$ Interesting question. Of course, one can compute the values $10-\pi^2$ for $p=3$ and $\frac1{45}\pi^4+\frac{10}3\pi^2-35$ for $p=4$ and probably, with more efforts, some more values... but to guess a general formula valid for every natural number $p\geqslant2$ is another matter. $\endgroup$
    – Did
    Nov 25, 2018 at 18:15
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    $\begingroup$ Yes, I would say the series is a valid answer. $\endgroup$
    – saulspatz
    Nov 25, 2018 at 18:23
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    $\begingroup$ You can also try to find a solution as a combination of polylogarithms. Write $\sum_{n=1}^{\infty} \frac{\lambda^{n+1}}{(n(n+1))^p} =$ some combination of polylogarithms of $\lambda$ and then let $\lambda=1$. $\endgroup$
    – mlerma54
    Nov 25, 2018 at 19:05
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    $\begingroup$ Ah you interested only in the case of integer $p$? $\endgroup$
    – J.G.
    Nov 25, 2018 at 20:09
  • $\begingroup$ Not just integer values, any real $p\geq1.$ $\endgroup$
    – Melody
    Nov 25, 2018 at 21:46

3 Answers 3


I was doing this exercise not a long ago with a student of mine, nice thing I have a chance to discuss it here. Assuming $p\in\mathbb{N}^+$, we just have to perform a partial fraction decomposition of $\frac{1}{x^p(x+1)^p}$. By stars and bars we have $\frac{1}{(1+x)^p}=\sum_{n\geq 0}\binom{p+n-1}{p-1}(-1)^n x^{n}$, hence the singularity at the origin gives a contribution equal to $$ \sum_{n=0}^{p-1}\binom{p+n-1}{p-1}\frac{(-1)^n}{x^{p-n}}=(-1)^p\sum_{n=1}^{p}\binom{2p-n-1}{p-1}\frac{(-1)^{n}}{x^{n}} $$ to the partial fraction decomposition of $\frac{1}{x^p(x+1)^p}$. By symmetry, the contribution provided by the singularity at $x=-1$ equals $$ (-1)^p\sum_{n=1}^{p}\binom{2p-n-1}{p-1}\frac{1}{(x+1)^n}$$ hence $\sum_{x\geq 1}\frac{1}{x^p(x+1)^p}$ splits as the following linear combination of values of the $\zeta$ function at even integers and telescopic series:

$$(-1)^p\sum_{k=1}^{\lfloor p/2\rfloor}\binom{2p-2k-1}{p-1}(2\zeta(2k)-1)-(-1)^p\sum_{k=0}^{\lfloor p/2\rfloor}\binom{2p-2k-2}{p-1} $$ which can be simplified as $$\boxed{\sum_{n\geq 1}\frac{1}{n^p(n+1)^p}= 2(-1)^p\sum_{k=1}^{\lfloor p/2 \rfloor}\binom{2p-2k-1}{p-1}\zeta(2k)-\frac{(-1)^p}{2}\binom{2p}{p}.}$$


$\newcommand{\msc}[2]{\left(\binom{#1}{#2}\right)}$ Assume that $p$ is a positive integer, and write $\msc{n}{k}$ for the multiset coefficient. Then by the method of Heaviside we have the partial fraction decomposition $$\frac{1}{n^p(n+1)^p}=(-1)^p\sum_{k=1}^{p}\msc{p}{p-k}\left(\frac{(-1)^k}{n^k}+\frac{1}{(1+n)^k}\right)$$ So that $$\begin{split}\sum_{n=1}^{\infty}\frac{1}{n^p(n+1)^p}&=(-1)^{p-1}\msc{p}{p-1} + (-1)^p\sum_{k=2}^p\msc{p}{p-k}(-1+(1+(-1)^k)\zeta(k))\\ &=(-1)^{p-1}\sum_{k=1}^p\msc{p}{p-k} + 2(-1)^p\sum_{k=1}^{\lfloor p/2\rfloor}\msc{p}{p-2k}\zeta(2k) \end{split}$$ $$\boxed{\sum_{n=1}^{\infty}\frac{1}{n^p(n+1)^p}=(-1)^{p-1}\sum_{k=1}^p\msc{p}{p-k} + 2(-1)^p\sum_{k=1}^{\lfloor p/2\rfloor}\msc{p}{p-2k}\frac{(2\pi)^{2k}\lvert B_{2k}\rvert }{2(2k)!}}$$ where $\zeta(s)$ denotes the Riemann zeta function and $B_n$ denotes the Bernoulli numbers.


Recurrence $$ \begin{align} \phi(a,b) &=\sum_{n=1}^\infty\frac1{n^a(n+1)^b}\\ &=\sum_{n=1}^\infty\left(\frac1{n^a(n+1)^{b-1}}-\frac1{n^{a-1}(n+1)^b}\right)\\[6pt] &=\phi(a,b-1)-\phi(a-1,b) \end{align} $$ where $\phi(a,0)=\zeta(a)$, $\phi(0,a)=\zeta(a)-1$, and $\phi(1,1)=1$.

Generating Function $$ \begin{align} \sum_{k=1}^\infty\sum_{n=1}^\infty\left(\frac{x(x+1)}{k(k+1)}\right)^n &=\frac{x(x+1)}{2x+1}\sum_{k=1}^\infty\frac{2x+1}{k(k+1)-x(x+1)}\\ &=-\frac{x(x+1)}{2x+1}\sum_{k=1}^\infty\left(\frac1{x-k}+\frac1{x+k+1}\right)\\ &=1-\frac{x(x+1)}{2x+1}\pi\cot(\pi x) \end{align} $$ Substituting $u=x(x+1)$, that is, $x=\frac{-1+\sqrt{1+4u}}2$, we get the generating function for $\sum\limits_{k=1}^\infty\frac1{k^n(k+1)^n}$ to be $$ 1+\frac{\pi u}{\sqrt{1+4u}}\tan\left(\frac\pi2\sqrt{1+4u}\right) $$


Start with $$ \frac1{n(n+1)}=\frac1n-\frac1{n+1} $$ then induction on $b$ shows that $$ \frac1{n(n+1)^b}=\frac1n-\sum_{k=1}^b\frac1{(n+1)^k} $$ and then induction on $a$ shows that $$ \frac1{n^a(n+1)^b} =\sum_{k=1}^a(-1)^{a-k}\frac{\binom{a+b-k-1}{a-k}}{n^k} +(-1)^a\sum_{k=1}^b\frac{\binom{a+b-k-1}{b-k}}{(n+1)^k} $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^a(n+1)^a}=(-1)^a\left[\sum_{k=1}^{\lfloor a/2\rfloor}2\binom{2a-2k-1}{a-1}\zeta(2k)-\binom{2a-1}{a-1}\right]} $$ Or since $\zeta(0)=-\frac12$, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^a(n+1)^a}=(-1)^a\sum_{k=0}^{\lfloor a/2\rfloor}2\binom{2a-2k-1}{a-1}\zeta(2k)} $$

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    $\begingroup$ You guys are really scary. $\endgroup$
    – Burrrrb
    Nov 27, 2018 at 4:18

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