I have encountered a problem in which you are asked to calculate the integral $\int_0^1x\ln(1+x)$ using a Riemann sum. The Riemann sum is $\sum_{k=1}^n\frac{k}{n^2}\ln(1+\frac{k}{n})$, and it should be evaluated as $n \rightarrow \infty$. I am stuck on how to proceed. I feel I need to find a closed expression for the sum, and then calculate the limit. My gut feeling is that this is somehow a geometric series in disguise, but I can find an expression for it. Any tips would be greatly appreciated!
Kind Regards, Andreas
We have that
$$\sum_{k=1}^n\frac{k}{n^2}\ln\left(1+\frac{k}{n}\right)=\sum_{k=1}^n\frac{k}{n^2}\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i}\left(\frac k n\right)^i=\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i}\sum_{k=1}^n\frac{k^{i+1}}{n^{i+2}}=\ldots$$
and since by Faulhaber's formula
$$\sum_{k=1}^n\frac{k^{i+1}}{n^{i+2}}=\frac1{i+2}+O\left(\frac1n\right)$$
we obtain
$$\ldots =\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i(i+2)}+ O\left(\frac1n\right)\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i}\to \frac14$$
indeed
$$O\left(\frac1n\right)\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i} = O\left(\frac1 n\right)\to 0$$
and
$$\sum_{i=1}^\infty \frac{(-1)^{i+1}}{i(i+2)} =\frac12\sum_{i=1}^\infty (-1)^{i+1}\left(\frac1{i}-\frac1{i+2}\right)=$$
$$\require{cancel}=\frac12\left[\left(1-\cancel{\frac13}\right)-\left(\frac12-\cancel{\frac14}\right)+\left(\cancel{\frac13}-\cancel{\frac15}\right)-\left(\cancel{\frac14}-\cancel{\frac16}\right)+\ldots \right]=\frac12\cdot \frac12 =\frac14$$