# 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.

• Maple says $\pi/2$ Commented Sep 4, 2012 at 14:20
• It looks like you might be able to interpret the summand as a Riemann sum ... Commented Sep 4, 2012 at 14:31
• @ Michael Joyce: Yeah. That's true. Anyway, I'm interested in more approaching ways if possible. Commented Sep 4, 2012 at 14:31

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.

• How is the decreasing function property called? Commented Nov 11, 2014 at 17:51
• @REr, decreasing Commented Nov 11, 2014 at 18:46

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 $\int_0^\infty f(x) dx$.

• With $a=n$, the approach becomes unconventional and seems to require some further analysis.
– Did
Commented Sep 4, 2012 at 17:42
• @did: I believe you're right. I've tried to add the necessary analysis to my answer.
– robjohn
Commented Sep 5, 2012 at 12:46

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 $$\lim_{n\to\infty}\sum_{k=0}^{n^2}\frac{n}{n^2+k^2}=\int_0^\infty\frac{\mathrm{d}x}{1+x^2}=\frac\pi2\tag{10}$$

• Nice and detailed (+1) Commented Sep 5, 2012 at 12:54
• @Chris'sister: This looks deceptively like a reference to Riemann Sums, but since the upper index is $n^2$, we have to handle the improper integral carefully.
– robjohn
Commented Sep 5, 2012 at 13:01
• @ robjohn: Right. And this is because m is not fixed, but it tends to $\infty$. Commented Sep 5, 2012 at 13:03

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 $$\begin{eqnarray*} \lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} &=& \lim_{n\to\infty}\left( \sum_{k=1}^\infty \frac{n}{n^2+k^2} - \sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2} \right) \\ &=& \lim_{n\to\infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2} \\ &=& \lim_{n\to\infty} \left(\frac{\pi}{2}\coth n\pi - \frac{1}{2n}\right) \\ &=& \frac{\pi}{2}. \end{eqnarray*}$$

• @Chris'sister: Sometimes life gets in the way! Aigner and Ziegler have a nice derivation of the partial fraction expansion for $\cot x$ in Proofs from The Book (see Cotangent and the Herglotz trick). I tend to think of such sums in terms of complex analysis. Commented Sep 6, 2012 at 16:43
• yeah, nice! Thanks for the link. Commented Sep 6, 2012 at 17:00

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\lim_{n \to \infty}\sum_{k = 1}^{n^{2}}{n \over n^{2} + k^{2}}:\ {\large ?}}$

\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}

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