Calculating $\sum_{n=1}^N \frac{1}{(N+1+n)(N+n)}$ by hand

Question

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?

Answer 1

Hint: $$\frac1{k(k+1)}=\frac1k-\frac1{k+1}.$$

Answer 2

Hint:$$\frac { 1 }{ (N+1+n)(N+n) } =\frac { 1 }{ N+n } -\frac { 1 }{ N+n+1 } $$

Answer 3

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