evaluate the sum $\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{(n^2+n-1)(k^2+k-1)}$ I'm trying to evaluate this sum 
$$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{(n^2+n-1)(k^2+k-1)}$$
I have no idea how to deal with it. 
With one sum I can, with partial-fraction decomposition, express it as a function of Digamma function and I stuck there.
 A: Suggesting too long for a comment:
When $f(n)$ is a positive function with $\sum_1^\infty f(n)$ converging: $$\left(\sum_{n=1}^\infty f(n)\right)^2=\sum_{n=1}^{\infty}\sum_{k=1}^\infty f(n)f(k)=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2$$
with $f(n) = \frac{1}{n^2+n-1}$
Then the sum you are looking for is:
$$\frac12\left(\left(\sum_{n=1}^\infty \frac{1}{n^2+n-1}\right)^2 +\sum_{n=1}^\infty \frac{1}{\left(n^2+n-1\right)^2}\right)$$
A: As already pointed by the other answers, we just have to compute:
$$ S_1=\sum_{n\geq 1}\frac{1}{n^2+n-1},\qquad S_2=\sum_{n\geq 1}\frac{1}{(n^2+n-1)^2}.\tag{1}$$
and by exploiting the logarithmic derivative of the Weierstrass product for the cosine function we have, for any $\alpha\in\mathbb{R}^+\setminus\mathbb{N}$:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-\alpha^2}=\frac{\pi\tan\frac{\pi\alpha}{2}}{4\alpha}\tag{2}$$
as well as:
$$ \sum_{n\geq 0}\frac{1}{\left((2n+1)^2-\alpha^2\right)^2}=\frac{\pi ^2 \sec^2\frac{\pi  \alpha }{2}}{16\, \alpha ^2}-\frac{\pi\tan\frac{\pi  \alpha }{2}}{8\, \alpha ^3}\tag{3}$$
hence it follows that:
$$ S_1 = 1+\frac{\pi}{\sqrt{5}}\,\tan\frac{\pi\sqrt{5}}{2}\tag{4} $$
and:
$$ S_2 = -1+\frac{\pi^2}{5}\,\sec^2\frac{\pi\sqrt{5}}{2}-\frac{2\pi}{5\sqrt{5}}\,\tan\frac{\pi\sqrt{5}}{2}\tag{5}$$
hence our double series equals:

$$\frac{\pi}{100}\,\sec^2\frac{\pi\sqrt{5}}{2}\left(15\pi-5\pi\cos(\pi\sqrt{5})+8\sqrt{5}\sin(\pi\sqrt{5})\right).\tag{6}$$

A: Let $p(x)=x^2+x-1$.
Then
$$\sum_{n=1}^\infty\sum_{k=n}^\infty\frac1{p(n)p(k)}=\frac12\left[\left(\sum_{n=1}^\infty\frac1{p(n)}\right)^2+\sum_{n=1}^\infty\frac1{p(n)^2}\right]$$
