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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