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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 $-\arctan{4}-\arctan{5} + \frac{\pi}{2}-\frac{\pi}{2}$ ?

2 Answers 2

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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}

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The telescoping partial sum gives $$\arctan(n+5)+\arctan(n+4)-\arctan(5)-\arctan(4)$$ hence the sum of the series is $$\pi-\arctan(5)-\arctan(4).$$

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    Why did you add the two $\frac{\pi}{2}$? Don't they cancel each other as we have the terms of our sum being $$\arctan{n+5} - \arctan{n+3} \to \lim_{n\to\infty}\arctan{n+5} -\lim_{n\to\infty}\arctan{n+3} \to \frac{\pi}{2}-\frac{\pi}{2}=0 $$? – 2017-02-24
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    Have a closer look to the simplification I got: there remains a *sum* of two arctangents. – 2017-02-24
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    Apologies for being dense, but I still don't see the limit of, $\arctan{(n+5)} - \arctan{(n+3)}$ being anything other than 0. We have the sum terms being in pairs, our last $n$ terms in the sum being $\ldots+(\arctan{(n+5)}-\arctan{(n+3)})$. Am I missing something? Edit: How exactly did you go about obtaining the sum of two terms which are positive? @FelixMarin used notation I have yet to become familiar with. – 2017-02-24