How to find $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$ find the limit value

$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}$$

this following is my methods:

let $$S_{n}=\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\sum_{j=1}^{n^2}\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\dfrac{1}{n}$$
  since
  $$\int_{\dfrac{j}{n}}^{\dfrac{j+1}{n}}\dfrac{dx}{1+x^2}<\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\cdot\dfrac{1}{n}<\int_{\dfrac{j-1}{n}}^{\dfrac{j}{n}}\dfrac{dx}{1+x^2}$$
  so
  $$\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}<S_{n}<\int_{0}^{n}\dfrac{dx}{1+x^2}$$

and note

$$\lim_{n\to\infty}\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{dx}{1+x^2}=\int_{0}^{infty}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$$
  so
  $$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\dfrac{\pi}{2}$$

I think this problem have other nice methods? Thank you 
and follow other methods 

$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{2}$$
  But there is a book say This methods is wrong,why, and where is wrong? Thank you 

 A: Another approach. Since:
$$ \int_{0}^{+\infty}\sin(ax)e^{-bx}\,dx = \frac{a}{a^2+b^2} \tag{1}$$
we have:
$$ \sum_{j=1}^{n^2}\frac{n}{n^2+j^2} = \int_{0}^{+\infty}\frac{1-e^{-n^2 x}}{e^x-1}\sin(nx)\,dx \tag{2}$$
where:
$$ \int_{0}^{+\infty}\frac{\sin(nx)}{e^x-1}\,dx = \text{Im}\int_{0}^{+\infty}\frac{e^{inx}}{e^x-1}\,dx=\sum_{k\geq 1}\frac{n}{n^2+k^2}=\frac{-1+\pi n \coth(\pi n)}{2n}\tag{3}$$
by Frullani's theorem and/or the logarithmic derivative of the Weierstrass product for the $\sinh$ function. It is quite trivial that the limit of the RHS of $(3)$ as $n\to +\infty$ is $\frac{\pi}{2}$, hence it is enough to prove that the contribute given by 
$$ \int_{0}^{+\infty}\frac{\sin(nx) e^{-n^2 x}}{e^x-1}\,dx\quad\text{or}\quad\sum_{k>n^2}\frac{n}{n^2+k^2}\tag{4}$$
is negligible. But that is quite easy.
A: Here is a method in the same spirit as what you have done. The main difference is that we explicitly estimate the difference between your sum and $\int_0^n\frac{{\rm d}x}{1+x^2}$.
Since $f$ is monotonely decreasing we have (see for example integral test on Wikipedia)
$$\left|\sum_{j=1}^n \frac{n}{n^2+(j+nm)^2} - \int_{m}^{m+1}\frac{{\rm d}x}{1+x^2}\right| < \frac{1}{n(1 + (m+1)^2)}$$
Using this we can estimate the difference between your sum and the integral of $\frac{1}{1+x^2}$ over $[0,n]$, which can be written $\sum_{m=0}^{n-1}\int_0^1\frac{{\rm d}x}{1+(x+m)^2}$, as
$$\left|\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} - \int_0^n\frac{{\rm d}x}{1+x^2}\right|  \leq \sum_{m=0}^{n-1} \left|\frac{1}{n}\sum_{j=1}^n \frac{1}{1+\left(\frac{j}{n}+m\right)^2} - \int_{0}^{1}\frac{{\rm d}x}{1+(x+m)^2}\right| \\\leq \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1+m^2} \leq \frac{1}{n}\sum_{m=1}^\infty \frac{1}{1+m^2}$$
This last sum can be evaluated as $\frac{\pi\coth(\pi)-1}{2}$, however all we need for this argument is that it's finite. This can be proved for example by comparison to $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}$. Taking $n\to\infty$ the result follows
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} = \lim_{n\to\infty}\int_0^n\frac{{\rm d}x}{1+x^2} = \frac{\pi}{2}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\sum_{j = 1}^{n^{2}}{n \over n^{2} + j^{2}} & =
-\,{1 \over n} + \Im\sum_{j = 0}^{n^{2}}{1 \over j - n\ic} =
-\,{1 \over n} +
\Im\sum_{j = 0}^{\infty}\pars{{1 \over j - n\ic} - {1 \over j + n^{2} - n\ic}}
\\[3mm] & =
-\,{1 \over n} +
\Im\sum_{j = 0}^{\infty}{n^{2} \over \pars{j + n^{2} - n\ic}\pars{j - n\ic}}
=-\,{1 \over n} + \Im\bracks{\Psi\pars{n^{2} - n\ic} - \Psi\pars{-n\ic}} 
\end{align}
where $\Psi$ is the Digamma function.
Then,
\begin{align}
\color{#f00}{\lim_{n \to \infty}\sum_{j = 1}^{n^{2}}{n \over n^{2} + j^{2}}} & =
\lim_{n \to \infty}\Im\bracks{\Psi\pars{n^{2} - n\ic} - \Psi\pars{-n\ic}}
\\[3mm] & =
\lim_{n \to \infty}\Im\braces{%
\bracks{\ln\pars{\root{n^{4} + n^{2}}} - \arctan\pars{{1 \over n}}\ic} -
\bracks{\vphantom{\Large A}\ln\pars{\verts{n}} - \arctan\pars{n}\ic}}
\\[3mm] &=
\lim_{n \to \infty}\arctan\pars{n} = \color{#f00}{{\pi \over 2}}
\end{align}

In using the asymptotic expansion of $\Psi\pars{z}$
  $\pars{~\mbox{when}\ z \to \infty~}$, we took care of the condition $\verts{\mathrm{arg}\pars{\ln\pars{z}}} < \pi^{-}$.

