Evaluating $\lim_{n\to \infty} \sum_{r=1}^{n}\frac{r}{n^2+n+r}$ The motive is to evaluate the following limit:
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{r}{n^2 + n + r}$$
I wrote it as 
$$ \lim_{n\to \infty}\sum_{r=1}^{n}\frac{r/n}{1 + 1/n + r/n^2} \approx ^{?} \int_{0} ^{1} x \, dx  = \frac{1}{2}$$
Now is this correct? Doesn't seem very correct to me. Thanks for your thoughts :)

Oliver Oloa gave a hint on Sandwich theorem but removed answer.
$$\sum_{r=1}^{n} \frac{r}{n^2 + n + n} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + 1}$$
Using this I think we get $1/2 \le L \le 1/2$ so limit is $1/2$.
 A: Oliver Oloa gave a hint on Sandwich theorem but removed answer.
$$\sum_{r=1}^{n} \frac{r}{n^2 + n + n} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + 1}$$
Using this we get:
$$ \frac{n(n+1)}{2(n^2 + n + n)} \le \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \le \frac{n(n+1)}{2(n^2 + n + 1)}$$
So as $n\to \infty$ the limit is $1/2$
A: Hint. One may also write
$$
\sum_{r=1}^{n} \frac{r}{n^2 + n + r}=(n^2+n)\left(H_{n^2+n}-H_{n^2+2n}\right)+n
$$ and conclude, as $n \to \infty$, with the asymptotics
$$
H_n=\ln n+\gamma+\frac1{2n}+O\left(\frac1{n^2}\right).
$$
A: The limit is $\frac{1}{2}$, since $\sum_{r=1}^{n}r = \frac{1}{2}n^2+\frac{1}{2}n$ and both $n^2+2n$ and $n^2+n+1$ are $n^2+O(n)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\lim_{n\to \infty}\sum_{r = 1}^{n}{r \over n^{2} + n + r} & =
\lim_{n\to \infty}\bracks{%
{1 \over n}\sum_{r = 1}^{n}{r \over n} +
\sum_{r = 1}^{n}\pars{{r \over n^{2} + n + r} - {r \over n^{2}}}}
\\[5mm] & =
\lim_{n\to \infty}\bracks{%
{1 \over n}\sum_{r = 1}^{n}{r \over n} -
\sum_{r = 1}^{n}{nr + r^{2} \over \pars{n^{2} + n + r}n^{2}}}
\end{align}

Note that

$$
0 < \sum_{r = 1}^{n}{nr + r^{2} \over \pars{n^{2} + n + r}n^{2}} <
\sum_{r = 1}^{n}{2n^{2} \over \pars{n^{2} + n + 1}n^{2}} = {2n \over n^{2} + n + 1}\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,{\large 0}
$$
such that
\begin{align}
\lim_{n\to \infty}\sum_{r = 1}^{n}{r \over n^{2} + n + r} & =
\lim_{n\to \infty}\pars{%
{1 \over n}\sum_{r = 1}^{n}{r \over n}} = \int_{0}^{1}x\,\dd x = \bbx{1 \over 2}
\end{align}
