If $n$ tends to infinity what will be the value of the sum $\arctan 1/3 +\arctan 1/7 +\arctan 1/13 +\arctan 1/21 +\cdots $ 
$$
\arctan\left(1 \over 3\right) + \arctan\left(1 \over 7\right) + \arctan\left(1 \over 13\right) + \arctan\left(1 \over 21\right) + \cdots 
$$
Estimate the value of the expression if $n \to \infty$, how to derive this and what will be the approach of this kind of question ?

I have tried to solve this by first doing partial sums then taking limit but I can't evaluate this after getting the form $\left(n^{2} + n + 1\right)$.
 A: $\arctan(x) - \arctan(y) = \arctan \left( \frac{x-y}{1+xy} \right)$
Now use $x = n+1$ and $y = n$ and you will get a telescoping sum.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{N \to \infty}\,\,\,\sum_{k = 0}^{N}\arctan\pars{1 \over k^{2} + 3k + 3} =
\lim_{N \to \infty}\,\,\,\sum_{k = 0}^{N}
\bracks{\arctan\pars{k + 2} - \arctan\pars{k + 1}}
\\[1cm] = &\
\lim_{N \to \infty}\left\lbrace\vphantom{\Large A}%
\bracks{\vphantom{\large A}\arctan\pars{2} - \arctan\pars{1}} + \bracks{\vphantom{\large A}\arctan\pars{3} - \arctan\pars{2}} + \cdots\right.
\\[5mm] &\ \left.\phantom{\lim_{N \to \infty}}%
\mbox{} + \bracks{\vphantom{\large A}\arctan\pars{N + 2} -
\arctan\pars{N + 1}}\right\rbrace
\\[8mm] = &\
\lim_{N \to \infty}\,\,\,
\bracks{\vphantom{\large A}-\arctan\pars{1} + \arctan\pars{N + 2}} =
-\,{\pi \over 4} + {\pi \over 2} = \bbx{\ds{\pi \over 4}}
\end{align}
