Question 15

We can write this sum as:

$\sum_{n=1}^\infty \frac{2n+1}{(n(n+1))^2} = \sum_{n=1}^\infty \frac{(n+1)^2-n^2}{n^2(n+1)^2} = \sum_{n=1}^\infty \frac1{n^2}-\frac1{(n+1)^2}.$

This sum telescopes (i.e. successive terms all cancel each other out), with only $\frac1{1^2}=1$ remaining. Hence the sum is $1$ , and the answer is $\boxed{\text{None of these}}$ .