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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

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    $\begingroup$ What you did is fine though, imo, a little too long and complicated. After the first equality you can pass directly to the integral since you know the integral (and the series, of course) exists and thus any Riemann sum does the job. $\endgroup$ – DonAntonio Sep 17 '13 at 11:09
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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.

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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

$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} = \lim_{n\to\infty}\int_0^n\frac{{\rm d}x}{1+x^2} = \frac{\pi}{2}$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\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 $\verts{\mathrm{arg}\pars{\ln\pars{z}}} < \pi^{-}$.

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