Telescoping Series Sum with arctan I see that this is a question which was previously asked. However, the answer which was given doesn't seem to shed any light on what a sums terms tend to.
$$\sum^{\infty}_{n=1}\left(\arctan{(n+5)-\arctan{(n+3)}}\right)$$
So far, I have that, the first $n$ terms are given by 
$$(\arctan{(6)} - \arctan{(4)})\qquad n=1\\
(\arctan{(7)} - \arctan{(5)})\qquad n=2\\
(\arctan{(8)} - \arctan{(6)})\qquad n=3\\
(\arctan{(9)} - \arctan{(7)})\qquad n=4\\
\vdots\\
(\arctan{(n+5)} - \arctan{(n+3)}) \qquad n=\infty\\\\$$
Furthermore, we have that we are left with $-\arctan{4}$ and $-\arctan{5}$ as these are the only terms, along with $\arctan{(n+5)} - \arctan{(n+3)}$ that do not cancel out.
Should our answer for this sum be $-\arctan{4}-\arctan{5} + \frac{\pi}{2}-\frac{\pi}{2}$ ?
 A: The telescoping partial sum gives
$$\arctan(n+5)+\arctan(n+4)-\arctan(5)-\arctan(4)$$
hence the sum of the series is
$$\pi-\arctan(5)-\arctan(4).$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{N \in \mathbb{N}_{\geq 1}}$:

\begin{align}
&\sum_{n = 1}^{N}\bracks{\arctan\pars{n + 5} - \arctan\pars{n + 3}} =
\sum_{n = 3}^{N + 2}\arctan\pars{n + 3} -
\sum_{n = 1}^{N}\arctan\pars{n + 3}
\\[1cm] = &
\bracks{\sum_{n = 3}^{N}\arctan\pars{n + 3} + \arctan\pars{N + 4} + \arctan\pars{N + 5}}
\\[5mm] - &\
\bracks{\arctan\pars{4} + \arctan\pars{5} + \sum_{n = 3}^{N}\arctan\pars{n + 3}}
\\[1cm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\, &\
\bbx{\ds{\pi - \arctan\pars{4} - \arctan\pars{5}}}
\end{align}
