Closed form of $\sum^{\infty}_{n=1} \dfrac{1}{n^a{(n+1)}^a}$ where $a$ is a positive integer
Let $h(z)= \frac1{(z+1)^k},g(z)=\frac{1}{z^k}$ then $$\frac{1}{z^k(z+1)^k} -\frac{\sum_{m=0}^{k-1} \frac{h^{(m)}(0)}{m!} z^m}{z^k}- \frac{\sum_{m=0}^{k-1} \frac{g^{(m)}(-1)}{m!} (z+1)^m}{(z+1)^k}$$
is a rational function with no pole thus it is a polynomial and since it vanishes at $\infty$ it is $0$.
$\frac{h^{(m)}(0)}{m!} =(-1)^m {m+k-1 \choose m}$, $\frac{g^{(m)}(-1)}{m!} = (-1)^{k}{m+k-1 \choose m}$
Thus $$\sum_{n=1}^\infty \frac{1}{n^k(n+1)^k}= \sum_{m=0}^{k-2}\zeta(k-m) {m+k-1 \choose m} ((-1)^m+ (-1)^{k})-\sum_{m=0}^{k-1} (-1)^{k}{m+k-1 \choose m}$$