The value of a real series What is the value of 
$$    \sum_{n=1}^\infty\frac{\sum_{k=1}^n \frac{1}{k^2}}{(n+1)(n+2)} $$
 A: First, change the order of summation (which we can do, since all terms are non-negative), to get
$$
\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{k^2}\frac{1}{(n+1)(n+2)}=\sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=k}^{\infty}\frac{1}{(n+1)(n+2)}.
$$
Using partial fractions on the inner sum, you'll find that this actually telescopes, yielding
$$
\sum_{n=k}^{\infty}\frac{1}{(n+1)(n+2)}=\frac{1}{k+1}.
$$
So, we are now trying to compute
$$
\sum_{k=1}^{\infty}\frac{1}{k^2(k+1)}.
$$
Another application of partial fractions yields
$$
\frac{1}{k^2(k+1)}=\frac{1}{k^2}-\frac{1}{k}+\frac{1}{k+1}.
$$
But
$$
\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6},
$$
and by again noticing telescoping we find that
$$
\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k}\right)=-1.
$$
Combining these then yields
$$
\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{k^2}\frac{1}{(n+1)(n+2)}=\frac{\pi^2}{6}-1.
$$
A: Your sum can be rewritten as
$$\sum_{n = 1}^{\infty}\bigg(\sum_{k = 1}^n {1 \over k^2}{1 \over (n+1)(n+2)}\bigg)$$
Changing the order of summation this becomes
$$\sum_{k = 1}^{\infty}\bigg(\sum_{n = k}^{\infty} {1 \over k^2}{1 \over (n+1)(n+2)}\bigg)$$
$$= \sum_{k = 1}^{\infty}{1 \over k^2}\bigg(\sum_{n = k}^{\infty}{1 \over (n+1)(n+2)}\bigg)$$
This telescopes into
$$\sum_{k = 1}^{\infty}{1 \over k^2(k+1)}$$
To sum this, use partial fractions... Key fact to be used is ${\displaystyle {1 \over k(k+1)} = {1 \over k} - {1 \over k+1}}$. So the sum is
$$\sum_{k = 1}^{\infty}{1 \over k}\bigg({1 \over k} - {1 \over k+1}\bigg)$$
$$= \sum_{k = 1}^{\infty}{1 \over k^2} - \sum_{k = 1}^{\infty}{1 \over k(k+1)} $$
First sum adds to ${\pi^2 \over 6}$, second telescopes to $1$, so the overall sum is 
$${\pi^2 \over 6} - 1$$
