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I have got to find the following limit:
$$\lim_{n\to\infty}\sum_{k=1}^{n} \arctan(\frac{1}{n+k})$$ At first I thought it was a Riemann sum, but I immediately got stuck at $\sum\limits_{k=1}^{n}\arctan(\frac{\frac{1}{n}}{1+\frac{k}{n}})$
Next I tried applying the transformation $\arctan(\frac{x-y}{1+xy})=\arctan(x)-\arctan(y)$ but I can't find the $x$ and $y$. Please help!

Edit: this is a contest-level question for the 11th grade in Romania, and I myself am in the 12th grade, so all these answers using Taylor series or Harmonic numbers are confusing to me. I was hoping for a high school level solution. Although thank you!

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Note that $\arctan(x)=x+O\!\left(x^3\right)$, then $$ \begin{align} \sum_{k=1}^n\arctan\left(\frac1{n+k}\right) &=\sum_{k=1}^n\left[\frac1{n+k}+O\!\left(\frac1{(n+k)^3}\right)\right]\\ &=\sum_{k=1}^n\frac1{1+k/n}\frac1n+O\left(\frac1{n^2}\right) \end{align} $$ Now it might be easier to use Riemann Sums.

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For small $x$ we have $$\arctan{x} \approx x$$ Which can be proven by examining the Taylor series of $\arctan{x}$. So as $n\to\infty$, the argument of the arctangent tends to $0$ and the limit is then equal to $$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n+k}=\lim_{n\to\infty} \sum_{i=n+1}^{2n} \frac{1}{i}=\lim_{n\to\infty} (H_{2n}-H_{n})$$ Where $H_n$ denotes the $n$th harmonic number given by $\sum_{k=1}^{n} \frac{1}{k}$. There is an asymptotic limit for the harmonic numbers that is given by $$H_n \approx \gamma + \ln{(n)}$$ for large $n$ where $\gamma$ denotes the Euler-Mascheroni constant. So our limit becomes $$\lim_{n\to\infty} (H_{2n}-H_{n})=\lim_{n\to\infty} (\gamma + \ln{(2n)}-(\gamma + \ln{(n)}))=\ln{(2)} $$

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    $\begingroup$ +1 I think this approach is right, but you need a little more control of $|\arctan(x) - x|$ in order to conclude (it's not enough just to note that it goes to zero, since you are adding up more and more terms of this form as $n \to \infty$.) $\endgroup$ – hunter Feb 23 at 14:28
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    $\begingroup$ If some wonder, $\mbox{arctan}(x) \simeq x$ can be made rigorous by using $|\mbox{arctan}(x)-x| \le \frac{|x|^3}{3}$ (can be shown by differentiating), and that $\frac{1}{3}\sum \limits_{k=1}^n \frac{1}{(n+k)^3} \underset{n\to \infty}{\longrightarrow}0$ $\endgroup$ – charmd Feb 23 at 14:32
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We have $$x-\frac{x^3}{3}<\tan^{-1}(x)<x,x>0$$

So $$\sum^{n}_{k=1}\bigg[\frac{1}{n+k}-\frac{1}{3(n+k)^3}\bigg]<\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)<\sum^{n}_{k=1}\frac{1}{n+k}$$

Now Using Limit $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n+k}=\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n}\cdot \frac{1}{1+\frac{k}{n}}$$

Using Riemann sum, $$ = \int^{1}_{0}\frac{1}{1+x}dx = \ln(2)$$

Same way

$$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{(n+k)^3}=\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n^2}\cdot \frac{1}{n}\cdot \frac{1}{(1+\frac{k}{n})^3}=0$$

And using Squeeze principle $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)=\ln(2)$$

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    $\begingroup$ How does this help in calculating the value of the sum? $\endgroup$ – Crostul Feb 23 at 14:18
  • $\begingroup$ From where do we have the first inequality? As I added in the post, I am only in high school so I don't know much about Taylor series or the like... $\endgroup$ – Constantin Feb 23 at 20:40
  • $\begingroup$ @Constantin: This is why we ask for context. If this is not mentioned in the question, then how can anyone know? It has been added now, but it is good to include things like this. $\endgroup$ – robjohn Feb 24 at 4:03
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As DXT did, using the upper and lower bounds of $\tan^{-1}(x)$ $$\sum^{n}_{k=1}\bigg[\frac{1}{n+k}-\frac{1}{3(n+k)^3}\bigg]<S_n=\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)<\sum^{n}_{k=1}\frac{1}{n+k}$$ $$\sum^{n}_{k=1}\frac{1}{n+k}=H_{2 n}-H_n$$ where appear harmonic numbers. $$\sum^{n}_{k=1}\frac{1}{(n+k)^3}=\frac{1}{2} (\psi ^{(2)}(2 n+1)-\psi ^{(2)}(n+1))$$ where appear polygamma functions.

Using the asymptotics, we then have $$\log (2)-\frac{1}{4 n}-\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right) \lt S_n < \log (2)-\frac{1}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right)$$

Using it for $n=10$, the left bound is $\log (2)-\frac{41}{1600}\approx 0.667522$, the right bound is $\log (2)-\frac{39}{1600}\approx 0.668772$ while the exact value is $\approx 0.667663$.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty} \sum_{k = 1}^{n}\arctan\pars{1 \over n + k}} \\[5mm] = &\ \lim_{n \to \infty} \sum_{k = 1}^{n}\pars{n + k} \int_{0}^{1}{\dd x \over x^{2} + \pars{n + k}^{2}} \\[5mm] = &\ \lim_{n \to \infty}\int_{0}^{1}\Re\sum_{k = 0}^{n - 1} {1 \over k + n + 1 - \ic x}\,\dd x \\[5mm] = &\ \lim_{n \to \infty}\Re\int_{0}^{1}\sum_{k = 0}^{\infty}\pars{% {1 \over k + n + 1 - \ic x} - {1 \over k + 2n + 1 - \ic x} }\,\dd x \\[5mm] = &\ \lim_{n \to \infty}\Re\int_{0}^{1}\bracks{% \Psi\pars{2n + 1 - \ic x} - \Psi\pars{n + 1 - \ic x}}\,\dd x \\[2mm] &\ \mbox{where}\ \pars{~\Psi:\ Digamma\ Function~} \\[5mm] = &\ -\lim_{n \to \infty}\Im \ln\pars{{\Gamma\pars{2n + 1 - \ic} \over \Gamma\pars{2n + 1}}\, {\Gamma\pars{n + 1} \over \Gamma\pars{n + 1 - \ic}}} \\[5mm] = &\ -\lim_{n \to \infty}\Im \ln\pars{\bracks{2n - \ic}!/\pars{2n}! \over \bracks{n - \ic}!/n!} = -\lim_{n \to \infty}\Im \ln\pars{\bracks{2n}^{-\ic} \over n^{-\ic}} \label{1}\tag{1} \\[5mm] = &\ -\Im\ln\pars{2^{-\ic}} = -\Im\ln\pars{\vphantom{\Large A}\cos\pars{\ln\pars{2}} - \ic\sin\pars{\ln\pars{2}}} \\[5mm] = &\ \bbx{\ln\pars{2}} \end{align}

$\ds{\Gamma}$ is the Gamma Function. Note that ( see \eqref{1} )

\begin{align} &\bbox[10px,#ffd]{\left.{\pars{\alpha n - \ic}! \over \pars{\alpha n}!}\,\right\vert_{\ \alpha\ \in\ \braces{1,2}}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {\root{2\pi}\pars{\alpha n - \ic}^{\alpha n - \ic + 1/2}\expo{-\pars{\alpha n - \ic}} \over \root{2\pi}\pars{\alpha n}^{\alpha n + 1/2}\expo{-\alpha n}} \\[5mm] = &\ {\pars{\alpha n}^{\alpha n - \ic + 1/2}\, \bracks{1 - \ic/\pars{\alpha n}}^{\alpha n - \ic + 1/2}\expo{\ic} \over \pars{\alpha n}^{\alpha n + 1/2}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {\large\pars{\alpha n}^{-\ic}} \end{align}

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