Evaluation of $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{r}{n^2+n+r}$$
$\bf{My\; Try:}$ Let $$L = \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{r}{n^2+n+r} = \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{\frac{r}{n}}{\frac{r^2}{n^2}+\frac{1}{n}+\frac{r}{n^2}}\cdot \frac{1}{n}$$
I want to convert into Riemann integral, But it is not possible here.
So how can I solve it?
Help me!
Thanks
$$\frac{r}{(n+1)^2}\leq\frac{r}{n^2+n+r}\leq\frac{r}{n^2}$$ hence your limit equals $\int_{0}^{1}x\,dx = \color{red}{\frac{1}{2}}$.
You may also avoid Riemann sums by just noticing that $\sum_{r=1}^{n} r = \frac{n^2+n}{2}$.
Besides Jack's neat answer, a different approach giving more than the desired limit.
One may rewrite your sum with the standard harmonic numbers $$ \begin{align} \sum^{n}_{r=1}\frac{r}{n^2+n+r}&=\sum^{n}_{r=1}\frac{n^2+n+r-(n^2+n)}{n^2+n+r}\\\\&=n-(n^2+n)\sum^{n}_{r=1}\frac1{n^2+n+r}\\\\ &=n-(n^2+n)\left(H_{n^2+2n}- H_{n^2+n}\right) \end{align} $$ then use the asymptotics of harmonic numbers, as $ N \to \infty$, $$ H_N=\log N+\gamma+\frac1{2N}-\frac1{12N^2}+\mathcal{O}\left(\frac1{N^4} \right) $$ leading readily to
$$ \sum^{n}_{r=1}\frac{r}{n^2+n+r}=\frac12-\frac1{3n}+\mathcal{O}\left(\frac1{n^2} \right) $$
as $n \to \infty$.
As an alternative, we have
$$\frac{r}{n^2+n+r}=\frac{r}{n^2+n}-\frac{r^2}{(n^2+n)(n^2+n+r)}$$
therefore
$$\sum^{n}_{r=1}\frac{r}{n^2+n+r}=\sum^{n}_{r=1}\frac{r}{n^2+n}-\sum^{n}_{r=1}\frac{r^2}{(n^2+n)(n^2+n+r)}=\frac12+O\left(\frac1n\right)\to\frac12$$