Calculating $\sum_{n=1}^N \frac{1}{(N+1+n)(N+n)}$ by hand In a recent proof I used induction to prove an identity concerning the harmonic progressions:
$$ \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\sum_{n=1}^{N}\frac{1}{N+n} $$ 
I needed to know what the following sum equaled so I used Wolfram Alpha to find, $N \in \mathbb{N}$:
$$ \sum_{n=1}^N \frac{1}{(N+1+n)(N+n)}=\frac{N}{2N^2+3N+1}$$ 
However, that got me thinking how you could go about finding that result without the help of a computer. 
Where would one even start? Try to manipulate it into a known format, where you already know a formula?

$$ \frac{1}{(N+n+1)(N+n)} = \frac{1}{N+n} - \frac{1}{N+n+1}, $$
and summing, the middle terms cancel and one is left with
$$ \sum_{n=1}^N \frac{1}{(N+n+1)(N+n)} = \frac{1}{N+1} - \frac{1}{2N+1} = \frac{N}{(2N+1)(N+1)}. $$
