Integral as a limit of sum: $\lim\limits_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)$ 
The value of$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)=\,?$$

Can the limit be partially applied to the denominator after converting the numerator into an integral?
I wrote this as $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\frac {k/n}{1+1/n+2k/n^2}\right)$$
From what I know I can write that $1/n$ as $\mathrm{d}x$ and $k/n$ as $x$. Can I apply limit to denominator as $n$ tends to infinity and rewrite the denominator as $1$.
 A: One the one hand
$$
\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}}  \le \mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2 }}}  = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  = \int_0^1 {x\, dx}  = \frac{1}{2}.
$$
On the other hand
\begin{align*}
\sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}}  \ge \sum\limits_{k = 1}^n {\frac{k}{{n^2  + 3n}}} & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{1 + \frac{3}{n}}}}  \ge \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\left( {1 - \frac{3}{n}} \right)} \\ & = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  - \frac{3}{n}\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} ,
\end{align*}
i.e.,
\begin{align*}
\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{k = 1}^n {\frac{k}{{n^2  + n + 2k}}} & \ge \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}}  - \mathop {\lim }\limits_{n \to  + \infty } \frac{3}{n}\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}} \\ & = \int_0^1 {x\, dx}  - 0 \cdot \int_0^1 {x\, dx}  = \frac{1}{2}.
\end{align*}
Thus, by the squeeze theorem, the limit is $\frac{1}{2}$.
A: Here's a hint: Use sandwich theorem. Put $k=n$ and $k=0$ in two cases and try to show your summation lies between these two. Then show that since limiting values of the other two summations is equal, this must have that value too.
A: What you can also do is to use generalied harmonic numbers since
$$S_n=\sum_{k=1}^n\left(\frac k{n^2+n+2k}\right)=\frac{1}{4} n \left((n+1) H_{\frac{1}{2} n (n+1)}-(n+1) H_{\frac{1}{2} n (n+3)}+2\right)$$
Now, using twice their asymptotics and continuing with Taylor series
$$S_n=\frac{1}{2}-\frac{2}{3 n}+\frac{4}{3
   n^2}+O\left(\frac{1}{n^3}\right)$$
