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.
 A: 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
$$\boxed{\sum_{n\geq 1}\frac{1}{n^p(n+1)^p}= 2(-1)^p\sum_{k=1}^{\lfloor p/2 \rfloor}\binom{2p-2k-1}{p-1}\zeta(2k)-\frac{(-1)^p}{2}\binom{2p}{p}.}$$
A: $\newcommand{\msc}[2]{\left(\binom{#1}{#2}\right)}$
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.
A: 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 $\zeta(0)=-\frac12$,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^a(n+1)^a}=(-1)^a\sum_{k=0}^{\lfloor a/2\rfloor}2\binom{2a-2k-1}{a-1}\zeta(2k)}
$$
