# Compute $\sum\limits_{n=1}^\infty\frac{1}{(n(n+1))^p}$ where $p\geq 1$

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.

• 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. – Did Nov 25 '18 at 18:15
• Yes, I would say the series is a valid answer. – saulspatz Nov 25 '18 at 18:23
• 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 $\lambda=1$. – mlerma54 Nov 25 '18 at 19:05
• Ah you interested only in the case of integer $p$? – J.G. Nov 25 '18 at 20:09
• Not just integer values, any real $p\geq1.$ – Melody Nov 25 '18 at 21:46

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 $$1+\frac{\pi u}{\sqrt{1+4u}}\tan\left(\frac\pi2\sqrt{1+4u}\right)$$

Induction

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 $$\zeta(0)=-\frac12$$, $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^a(n+1)^a}=(-1)^a\sum_{k=0}^{\lfloor a/2\rfloor}2\binom{2a-2k-1}{a-1}\zeta(2k)}$$

• You guys are really scary. – MereMortal47 Nov 27 '18 at 4:18