<s>[INST] Evaluate $\lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$ Evaluate 

$$  \lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$$

I used definite integral as a limit of a sum as:
$$S= \lim_{ n \to \infty}\frac{1}{n} \sum_{r=1}^{n} \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2}$$
So
$$S=\int_{0}^{1}\frac{ x \:dx}{1+x^2}=\frac{1}{2} \log 2$$
Is there any other approach?
 [/INST] Using a double limit with uniform convergence,
$$\begin{align}\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2 + k } &= \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k / n^2} \\&=\lim_{m \to \infty}\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k /(nm)} \\ &= \lim_{m \to \infty} \int_0^1 \frac{x}{1 + x/m} \, dx \\ &= \int_0^1 x \, dx  \\ &= \frac{1}{2}\end{align}$$
Note that if $a_{nm} \to b_n$ uniformly, then 
$$\lim_{n\to \infty}a_{nn} = \lim_{n\to \infty} \lim_{m \to \infty}a_{nm} = \lim_{m\to \infty} \lim_{n \to \infty}a_{nm}$$
</s>