Define $$f(x)= \begin{cases}2rx,&\text{if }\frac{1}{r+1}<x<=\frac{1}{r}; r=1,2,...\\ 0,&\text{if } x=0. \end{cases}$$ Prove that $f$ is Riemann integrable in [0,1] and find integral $f$ from $0$ to $1$.
Since $f$ has countably infinite points of discontinuity but has only one limit point $0$, it is Riemann integrable. Now, here is the confusion, I tried to find the value of integration using upper darboux sum and got answer as $2$ as l.u.b of all the partitions will be $2$ and $U(f,p)=2-1/(n+1)$. Hence after applying limit answer should be $2$ but it is incorrect. Where have I made mistake?
Consider this: \begin{align*} \int_0^1 f(x)\,dx = \sum_{r=1}^\infty\int_{1/(r+1)}^{1/r} f(x)\,dx &= \sum_{r=1}^\infty\int_{1/(r+1)}^{1/r} 2rx\,dx = \sum_{r=1}^\infty \frac{2r+1}{r(r+1)^2}\\ &= \sum_{r=1}^\infty \left(\frac1{(r+1)^2} + \frac1r - \frac1{r+1}\right). \end{align*} Now, it's well known that $\sum\limits_{n=1}^\infty \dfrac1{n^2} = \dfrac{\pi^2}6$. Can you finish?