$$\lim_{n\to\infty}\sum\limits_{0\leqslant{k}\leqslant{2n}} {\frac{k}{k+n^{2}}}$$
I can't figure out the "right" function for this limit. The previous problem was very similar except there was $k^2$ in the denominator, and it wasn't difficult to recognize the Riemann sum for $f(x)=\frac{x}{x^2+1}$.
Given the limits, this integral sum can be written as $\sum{f(\frac{2k}{n})\cdot\frac{2}{n}}$.
On
Write the sum as
$$S_{nn} = \frac{1}{n}\sum_{k=0}^{2n} \frac{k/n}{1 + k/n^2}$$
This is not in the form of a Riemann sum. However, there is a trick (requiring justification) where we can reduce the limit to that of a Riemann sum by viewing this in terms of a double sequence
$$S_{nm} = \frac{1}{n}\sum_{k=0}^{2n} \frac{k/n}{1 + (k/n)(1/m)}$$
Clearly, $\displaystyle\lim_{m \to \infty} S_{nm} = \frac{1}{n}\sum_{k=0}^{2n} \frac{k}{n}.\,$ As the convergence can be shown to be uniform for all $n$, by a well-known theorem for double sequences, we can evaluate as an iterated limit,
$$\lim_{n \to \infty} S_{nn} = \lim_{n \to \infty} \lim_{m \to \infty} S_{nm} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=0}^{2n} \frac{k}{n} = \int_0^2 x \, dx = 2.$$
Justification: Uniform convergence of inner limit
Note that
$$\left|\frac{1}{n}\sum_{k=0}^{2n} \frac{k/n}{1 + (k/n)(1/m)} - \frac{1}{n}\sum_{k=0}^{2n} \frac{k}{n} \right|= \frac{1}{n}\sum_{k=0}^{2n}\frac{(k/n)^2(1/m)}{1 + (k/n)(1/m)} \\ \leqslant \frac{1}{nm}\sum_{k=0}^{2n}\left(\frac{k}{n}\right)^2 \leqslant \frac{2n }{nm}\left(\frac{2n}{n} \right)^2 = \frac{8}{m}$$
Edited after reading the comments.
No need for Riemann sums.
Lower bound: $$ \sum_{k=0}^{2n}\frac{k}{k+n^2}\ge\frac{1}{2\,n+n^2}\sum_{k=0}^{2n}k=\frac{n\,(2\,n+1)}{2\,n+n^2}. $$ Upper boud: $$ \sum_{k=0}^{2n}\frac{k}{k+n^2}\le\frac{1}{n^2}\sum_{k=0}^{2n}k=\frac{n\,(2\,n+1)}{n^2}. $$ However, Riemann sums can be used, but in a more complicated way: $$ \sum_{k=0}^{2n}\frac{k}{k+n^2}=2\,n\,\Bigl(\frac{1}{2\,n}\sum_{k=0}^{2n}\frac{k/(2\,n)}{k/(2\,n)+n/2}\Bigr) \sim2\,n\int_0^1\frac{x\,dx}{x+n/2}. $$