How does one determine that $\sum_{k=1}^{n} \frac{1}{(n+k)(n+k+1)}$ evaluates to $\frac{n}{2n^2+3n+1}$?

Question

How does one determine that $$\sum_{k=1}^{n} \frac{1}{(n+k)(n+k+1)}$$ evaluates to $$\frac{n}{2n^2+3n+1}\ ?$$ What are the simplification steps involved?

Answer 1

The idea is to note that:

$$\frac{1}{(n+k)(n+k+1)} = \frac{n+k+1-(n+k)}{(n+k)(n+k+1)} = \frac{1}{n+k} - \frac{1}{n+k+1}$$

So your sum telescopes to:

$$\sum_{k=1}^{n} \frac{1}{(n+k)(n+k+1)} = \sum_{k=1}^n\left(\frac{1}{n+k} - \frac{1}{n+k+1}\right)$$ $$= \frac{1}{n+1}-\frac{1}{n+n+1} = \frac{n}{2n^2+3n+1}$$