TITLE: Evaluation of $\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{r}{n^2+n+r}$
QUESTION [5 upvotes]: 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 Reinmann Integral, But it is not possible here.
So How can I solve it
Help me
Thanks

REPLY [4 votes]: Besides Jack's neat answer, a different approach giving more than the desired limit. 
One may rewrite your sum with the standard harmonic numbers 
$$
\begin{align}
\sum^{n}_{r=1}\frac{r}{n^2+n+r}&=\sum^{n}_{r=1}\frac{n^2+n+r-(n^2+n)}{n^2+n+r}\\\\&=n-(n^2+n)\sum^{n}_{r=1}\frac1{n^2+n+r}\\\\
&=n-(n^2+n)\left(H_{n^2+2n}- H_{n^2+n+1}\right)
\end{align}
$$ 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 $n \to \infty$.