Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ How would you evaluate the following series?
$$\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $$
Thanks.
 A: Hint: $$\int_0^a f(x) dx \approx \sum_{k=0}^{na} \frac{1}{n}f\left(\frac{k}{n}\right)$$ Use $f(x) = \frac{1}{1+x^2}$.
Addendum:
Fortunately, $f(x)$ is strictly decreasing, therefore the error is bounded by $\frac{f(0)-f(a)}n$, which again is $<\frac1n$, independent of $a$. This last observation allows us to use $a=n$ without spoiling convergence to $\int_0^\infty f(x) dx$.
A: Using $x=k/n$ and $\mathrm{d}x=1/n$
$$
S_m(n)=\sum_{k=0}^{mn}\frac{n}{n^2+k^2}=\sum_{k=0}^{mn}\frac{1}{1+(k/n)^2}\frac1{\vphantom{k^2}n}\tag{1}
$$
is a Riemann Sum for
$$
I_m=\int_0^m\frac{\mathrm{d}x}{1+x^2}\tag{2}
$$
For any $m$ and $n$, we have
$$
\sum_{k>mn}\frac{n}{n^2+k^2}\le\sum_{k>mn}\frac{n}{k(k-1)}=\frac1m\tag{3}
$$
which implies that
$$
S_m(n)\le S_\infty(n)=\lim_{m\to\infty}S_m(n)\le S_m(n)+\frac1m\tag{4}
$$
Since
$$
I_\infty=\lim\limits_{m\to\infty}I_m=\int_0^\infty\frac{\mathrm{d}x}{1+x^2}\tag{5}
$$
for any $\epsilon>0$, there is an $m_\epsilon\ge\frac1{\large\epsilon}$ so that for $m\ge m_\epsilon$,
$$
I_\infty-\epsilon\le I_m=\lim_{n\to\infty}S_m(n)\le I_\infty\tag{6}
$$
Finally, there is an $n_\epsilon\ge m_\epsilon$ so that for $n\ge n_\epsilon$,
$$
I_\infty-2\epsilon\le S_{m_\epsilon}(n)\le I_\infty+\epsilon\tag{7}
$$
Since $m_\epsilon\ge\frac1{\large\epsilon}$, $(4)$ and $(7)$ yield that for $n\ge n_\epsilon$
$$
I_\infty-3\epsilon\le S_n(n)\le I_\infty+2\epsilon\tag{8}
$$
Since $\epsilon$ was arbitrary, we get that
$$
\lim_{n\to\infty}S_n(n)=I_\infty\tag{9}
$$
which translates to
$$
\lim_{n\to\infty}\sum_{k=0}^{n^2}\frac{n}{n^2+k^2}=\int_0^\infty\frac{\mathrm{d}x}{1+x^2}=\frac\pi2\tag{10}
$$
A: Here's another approach.
First, note that
$$\begin{eqnarray*}
\sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2}
&<& \sum_{k=n^2+1}^\infty \frac{n}{k^2} \\
&\le& n\int_{n^2}^\infty \frac{dx}{x^2} \\
&=& \frac{1}{n}.
\end{eqnarray*}$$
We also need the partial fraction expansion of $\coth x$,
$$\begin{eqnarray*}
\coth x &=& \lim_{N\to\infty} \sum_{k=-N}^N \frac{1}{x-i k \pi} \\
&=& \frac{1}{x} + \sum_{k=1}^\infty \frac{2x}{x^2+k^2\pi^2}.
\end{eqnarray*}$$
Then we find 
$$\begin{eqnarray*}
\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2}
&=& \lim_{n\to\infty}\left(
\sum_{k=1}^\infty \frac{n}{n^2+k^2} - \sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2}
\right) \\
&=& \lim_{n\to\infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2} \\
&=& \lim_{n\to\infty} \left(\frac{\pi}{2}\coth n\pi - \frac{1}{2n}\right) \\
&=& \frac{\pi}{2}.
\end{eqnarray*}$$
A: Recall that for any decreasing function $f:\mathbb{R}\to\mathbb{R}$ and any $N>1$ we have
$$
\int\limits_1^{N+1}f(x)dx\leq \sum\limits_{k=1}^{N}f(k)\leq \int\limits_0^N f(x)dx
$$
After substitutions $N=n^2$, $f(x)=n/(n^2+x^2)$ and simple computations we have
$$
\arctan\frac{n^2+1}{n}-\arctan \frac{1}{n}\leq\sum\limits_{k=1}^{n^2}\frac{n}{n^2+k^2}\leq\arctan n
$$
Lets take a limit $n\to\infty$, then from sandwich lemma it follows
$$
\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n^2}\frac{n}{n^2+k^2}=\frac{\pi}{2}
$$
P.S. First solution was not rigor enough.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\lim_{n \to \infty}\sum_{k = 1}^{n^{2}}{n \over n^{2} + k^{2}}:\ {\large ?}}$

\begin{align}
\color{#c00000}{\sum_{k = 1}^{n^{2}}{n \over n^{2} + k^{2}}}&=
-\Im\sum_{k = 1}^{n^{2}}{1 \over k + \ic n}
=-\Im\sum_{k = 0}^{n^{2} - 1}{1 \over k + 1 + \ic n}
\\[3mm]&=-\Im\sum_{k = 0}^{\infty}\pars{%
{1 \over k + 1 + \ic n} - {1 \over k + n^{2} + 1 + \ic n}}
\\[3mm]&=-n^{2}\,\Im\sum_{k = 0}^{\infty}
{1 \over \pars{k + n^{2} + 1 + \ic n}\pars{k + 1 + \ic n}}
\\[3mm]&=\color{#c00000}{%
\Im\bracks{\Psi\pars{1 + \ic n} - \Psi\pars{n^{2} + 1 + \ic n}}}
\end{align}
  where $\ds{\Psi\pars{z}}$ is the
  Digamma Function.
  $\ds{z \in {\mathbb C}\verb=\=\braces{0,-1,-2,\ldots}}$. We'll use the property $\ds{\Psi\pars{z} \approx \ln\pars{z}}$ when $\ds{\verts{z} \gg 1}$
  and $\ds{\verts{{\rm Arg}\pars{z}} < \pi}$.

Then,
\begin{align}\color{#00f}{\large%
\lim_{n \to \infty}\sum_{k = 1}^{n^{2}}{n \over n^{2} + k^{2}}}&=
\lim_{n \to \infty}\Im\bracks{\Psi\pars{1 + \ic n} - \Psi\pars{n^{2} + 1 + \ic n}}
\\[3mm]&=\lim_{n \to \infty}
\Im\bracks{\ln\pars{1 + \ic n} - \ln\pars{n^{2} + 1 + \ic n}}
\\[3mm]&=
\lim_{n \to \infty}\bracks{\arctan\pars{n} - \arctan\pars{n \over n^{2} + 1}}
=\color{#00f}{\large{\pi \over 2}}
\end{align}
A: We have the following important theorem,

Theorem: Let for the monotonic function $f$ ,$\int_{0}^\infty f(x)dx$ exists and we have $\lim_{x\to\infty}f(x)=0$ and $f(x)>0$ then
  we have 
$$\lim_{h\to0^+}h\sum_{v=0}^\infty f(vh)=\int_{0}^\infty f(x)dx$$

It is enough to take $h^{-1}=t$ and $f(x)=\frac{2}{1+x^2}$, then we get the desired result.

So we showed that
  $$\lim_{t\to\infty}\left(\frac{2}{t}+\frac{2t}{t^2+2^2}+\cdots+\frac{2t}{t^2+n^2}+\cdots\right)=\pi$$

Let me show some additional infinite sum by this theorem

We show 
$$\lim_{t\to
 1^{-}}(1-t)^2\left(\frac{t}{1-t}+\frac{2t^2}{1-t^2}+\frac{3t^2}{1-t^2}+\cdots
 \frac{nt^2}{1-t^2}+\cdots\right)=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$

Now in previous theorem if we take $e^{-h}=t$ and $f(x)=\frac{xe^{-x}}{1-e^{-x}}$ then since
$$\int_0^\infty\frac{xe^{-x}}{1-e^{-x}}=\int_0^\infty x(\sum_{n=1}^\infty e^{-nx})dx=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
The previous theorem gives us an important relation of infinite sum about Euler constant

We show
$$\lim_{t\to
 1^{-}}(1-t)\left(\frac{t}{1-t}+\frac{t^2}{1-t^2}+\frac{t^3}{1-t^3}+\cdots \frac{t^n}{1-t^n}+\cdots\right)-\log\frac{1}{1-t}=C=\text{Euler
 constant}$$

By using the previous theorem we just need to take $f(x)=\frac{e^{-x}}{1+e^{-x}}$ and $e^{-h}=t$ and since it is known that $$\int_0^\infty e^{-x}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)dx=C=\text{Euler constant}$$
See this book for more examples of G.polya, problems and theorems in analysis , vol  I  Springer 
