The current American Mathematical Monthly has the identity
$\frac1{(n(n+1))^3} =6(\frac1{n}-\frac1{n+1}) -3(\frac1{n^2}+\frac1{(n+1)^2}) +(\frac1{n^3}-\frac1{(n+1)^3}) $
which is used to show that $10-\pi^2 =\sum_{n=1}^{\infty} \frac1{(n(n+1))^3} $.
This led me to ask:
What about $\frac1{(n(n+1))^m} $?
Experimentation led to this conjecture:
$\frac1{(n(n+1))^m} =\sum_{k=0}^{m-1} a_{m,k}(-1)^k(\frac1{n^{m-k}}+(-1)^{m-k}\frac1{(n+1)^{m-k}}) $ where $a_{m, k} =\binom{ m-1+k}{k} =\binom{ m-1+k}{m-1},\\ $
and this is what I would like a proof of.
This has been verified for $1 \le m \le 5$.
Note that this leads to the following results:
If $$s(m) =\sum_{n=1}^{\infty} \frac1{(n(n+1))^m} $$ then $$s(m) =(-1)^m2\sum_{k=1}^{\lfloor m/2 \rfloor} a_{m,m-2k}\zeta(2k)-(-1)^m\sum_{k=0}^{m-1} a_{m,k} $$ so that $$s(2m) =2a_{2m,0}\zeta(2m)+2\sum_{k=1}^{m-1} a_{2m,2m-2k}\zeta(2k)-\sum_{k=0}^{2m-1} a_{2m,k} $$ and $$s(2m+1) =-2a_{2m+1,1}\zeta(2m)-2\sum_{k=1}^{m-1} a_{2m+1,2m-2k+1}\zeta(2k) +\sum_{k=0}^{2m} a_{2m+1,k} $$
Too bad this shows nothing about zeta of odd parameter.