Evaluate $\lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$ Evaluate 

$$  \lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$$

I used definite integral as a limit of a sum as:
$$S= \lim_{ n \to \infty}\frac{1}{n} \sum_{r=1}^{n} \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2}$$
So
$$S=\int_{0}^{1}\frac{ x \:dx}{1+x^2}=\frac{1}{2} \log 2$$
Is there any other approach?
 A: Using a double limit with uniform convergence,
$$\begin{align}\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2 + k } &= \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k / n^2} \\&=\lim_{m \to \infty}\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k /(nm)} \\ &= \lim_{m \to \infty} \int_0^1 \frac{x}{1 + x/m} \, dx \\ &= \int_0^1 x \, dx  \\ &= \frac{1}{2}\end{align}$$
Note that if $a_{nm} \to b_n$ uniformly, then 
$$\lim_{n\to \infty}a_{nn} = \lim_{n\to \infty} \lim_{m \to \infty}a_{nm} = \lim_{m\to \infty} \lim_{n \to \infty}a_{nm}$$
A: HINT:
$$\frac{n(n+1)}{2}\frac{1}{n+n^2}\le \sum_{k=1}^n \frac{k}{k+n^2}\le \frac{n(n+1)}{2}\frac{1}{1+n^2}$$
A: $$ \sum_{k=1}^{n}\frac{k}{k+n^2} = n-n^2\sum_{k=1}^{n}\frac{1}{k+n^2} = n-n^2\left(H_{n^2+n}-H_{n^2}\right) $$
and since $H_m = \log(m)+\gamma+\frac{1}{2m}+O\left(\frac{1}{m^2}\right) $ we have
$$ \lim_{n\to +\infty}\sum_{k=1}^{n}\frac{k}{k+n^2} = \lim_{n\to +\infty}\left[n-n^2\log\left(1+\frac{1}{n}\right)\right]=\color{red}{\frac{1}{2}}. $$

An alternative approach is provided by summation by parts:
$$\sum_{k=1}^{n}\frac{k}{k+n^2} = \frac{1}{2}+\sum_{k=1}^{n-1}\frac{k(k+1)}{2}\left(\frac{1}{k+n^2}-\frac{1}{k+1+n^2}\right)$$
where the last sum is bounded by
$$ \sum_{k=1}^{n-1}\frac{n^2}{2n^4}\leq\frac{1}{2n}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\underline{\textsf{Original}}}$ OP question:

  $\ds{\lim_{n \to \infty}\pars{{1 \over 1^{2} + n^{2}} + {2 \over 2^{2} + n^{2}} + {3 \over 3^{2} + n^{2}} + \cdots +
{n \over n^{2} + n^{2}}} =
\lim_{n \to \infty}\sum_{k = 1}^{n}{k \over k^{2} + n^{2}}:\ {\large ?}}$.

\begin{align}
\lim_{n \to \infty}\sum_{k = 1}^{n}{k \over k^{2} + n^{2}} & =
\lim_{n \to \infty}\Re\sum_{k = 0}^{n - 1}{1 \over k + 1 + \ic n} =
\lim_{n \to \infty}\Re\sum_{k = 0}^{\infty}\pars{%
{1 \over k + 1 + \ic n} - {1 \over k + n + 1 + \ic n}}
\\[5mm] & =
\lim_{n \to \infty}\Re\sum_{k = 0}^{\infty}\pars{H_{n + \ic n} - H_{\ic n}}
\qquad \pars{~H_{z}:\ Harmonic\ Number~}
\end{align}

Note that
  $\ds{H_{z} \sim \ln\pars{z} + \gamma + {1 \over 2z} - {1 \over 12z^{2}} +
\,\mrm{O}\pars{1 \over z^{4}}}$ as $\ds{\verts{z} \to \infty}$
  where $\ds{\gamma}$ is the Euler-Mascheroni Constant.


Then, 
\begin{align}
\lim_{n \to \infty}\sum_{k = 1}^{n}{k \over k^{2} + n^{2}} & =
\lim_{n \to \infty}\Re\ln\pars{n + \ic n \over \ic n} =
\Re\ln\pars{1 - \ic} = \ln\pars{\root{1^{2} + 1^{2}}} = \bbx{{1 \over 2}\,\ln\pars{2}}
\end{align}
