Integral of complicated simple function

Question

In my homework I am trying to find the integral of:

$u(x)=\sum_{n=1}^\infty \frac{1}{n^2(n+1)}*1_[0,n]$

Using Excel, I can calculate that the integral is 1. However, when I try to show this mathematically, I end up with an expression I cannot find the sum of.

I get for $x \in [0,1]$: $u(x)=\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n(n+1)}$

and for $x \in (1,2]$: $u(x)=\sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n(n+1)} - \frac{1}{1^2(1+1)}$

But this is as far as I go, any help would be appreciated

Answer 1

$\int_0^\infty \sum_{n=1}^\infty \frac{1}{n^2(n+1)}1_{[0,n]} dx=$

$=\sum_{n=1}^\infty\int_0^\infty \frac{1}{n^2(n+1)}1_{[0,n]}dx=$

$=\sum_{n=1}^\infty\int_0^n \frac{1}{n^2(n+1)}dx=$

$=\sum_{n=1}^\infty \frac{1}{n^2(n+1)}\int_0^ndx=$

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

$=\sum_{n=1}^\infty \frac{1}{n(n+1)}=$

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

$=\sum_{n=1}^\infty \frac{1}{n}-\frac{1}{n+1}=$

$=\lim_{k\to\infty}\sum_{n=1}^k \frac{1}{n}-\frac{1}{n+1}=$

$=\lim_{k\to\infty} 1-\frac{1}{k+1}=1$

You can observe that $\sum_{n=1}^\infty \frac{1}{n}-\frac{1}{n+1}$ is a telescopic series.