Evaluate $$ \lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k}{n^2+k}\ln(1+\frac{k}{n}). $$
I think it may be associated with the definition of Riemann integration. Since we know that $$ \lim_{n\to+\infty}\sum_{k=1}^{n}\frac{1}{n}\ln(1+\frac{k}{n})=\int_0^1\ln(1+x)\,dx, $$ what I want to do is like this.
This is not in the form of a Riemann sum, however we can evaluate as
$$\begin{align}\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2+k} \ln(1 + k/n) &= \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{k/n}{1+k/n^2} \ln(1 + k/n)\\ &= \lim_{n \to \infty} \lim_{m \to \infty} \underbrace{\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1+(k/n)(1/m)} \ln(1 + k/n)}_{\text{uniformly convergent for all } n \in \mathbb{N}} \\ &=\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n (k/n) \ln(1 + k/n) \\ &= \int_0^1 x \ln (1+x) \, dx \end{align}$$
Evaluation as a double limit is justified by the uniform convergence of the inner limit on the second line.
The uniform convergence follows from
$$\left|\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1+(k/n)(1/m)} \ln(1 + k/n) - \sum_{k=1}^n (k/n) \ln(1 + k/n) \right| \\ \leqslant \frac{1}{n}\sum_{k=1}^n \frac{(k/n)^2(1/m)}{1 + (k/n)(1/m)} \ln(1 + k/n) \\ \leqslant \frac{1}{nm}\sum_{k=1}^n (k/n)^2 \ln (1 +k/n)\\ \leqslant \frac{\ln(2)}{m}$$
