Calculating $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n} \arctan(\frac{1}{n+k})$ 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!
 A: 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.
A: 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
$$\lim_{n\to\infty} (H_{2n}-H_{n})=\lim_{n\to\infty} (\gamma + \ln{(2n)}-(\gamma + \ln{(n)}))=\ln{(2)} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: 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 $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)=\ln(2)$$
A: 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 $\approx 0.667663$.
