Problem on Riemann Integration: $f$ is defined on $[0,1]$ by $\,f(x)=1/n$ for $1/n>x\geq 1/(n+1)$ and $f(x)=0$ for $x=0$, where $n=1,2,3,....$
Find $\int_{0}^1f(x)dx$.
First I tried to find out $\int_{1/(n+1)}^1f(x)dx$. But I got a series sum:$$\sum_{k=1}^{n}\cfrac{1}{k^2(k+1)}$$
How to proceed?
since $(0,1) = \bigcup_{n=1}^{\infty} ( \frac1{n+1} , \frac1{n })$. (pairwise disjoint union)
$$\int_{0}^1f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}f(x)dx = \sum_{n=1}^{ \infty} \int_{1/n+1}^{1/n}\frac1ndx=\sum_{n=1}^{ \infty} \frac1{n^2 }- \frac1{(n+1)n}= \frac{\pi^2}{6}-1 $$
Given that $ \frac1{(n+1)n}= \frac1{n}- \frac1{n+1}$ then $$\sum_{n=1}^{ \infty} \frac1{(n+1)n}=\sum_{n=1}^{ \infty} (\frac1{n}- \frac1{n+1})=1$$
