Say we have the following series $\displaystyle \sum_{n=5}^\infty (\arctan(n^{-1})-\arctan((n+1)^{-1}))$
I want to determine whether it is convergent or divergent, here's what I tried:
$\displaystyle \sum_{n=5}^\infty (\arctan(n^{-1})-\arctan((n+1)^{-1}))$
$= \displaystyle \lim_{n\to\infty}\displaystyle \sum_{k=5}^n (\arctan(k^{-1})-\arctan((k+1)^{-1}))$
which is a telescoping series that can be simplified to
$= \displaystyle \lim_{n\to\infty}(\arctan(5^{-1})-\arctan((n+1)^{-1}))$
= $\arctan(5^{-1})$
And so the series converges to $\arctan(5^{-1})$. Is that correct?
Yes. It might even be more obvious if you write out a few terms. Then
$$\begin{align} \sum_{n=5}^\infty \arctan \left( \frac 1 n \right) - \arctan \left( \frac{1}{n+1} \right) = &\;\;\;\; \arctan \left( \frac 1 5 \right) - \arctan \left( \frac 1 6 \right) \\ &+ \arctan \left( \frac 1 6 \right) - \arctan \left( \frac 1 7 \right) \\ &+ \arctan \left( \frac 1 7 \right) - \arctan \left( \frac 1 8 \right) \\ &+ \cdots\\ \end{align}$$
Obvious cancellation happens along the diagonals, leaving us with $\arctan(1/5)$ ... on the premise it converges! Bear in mind that, for this telescoping method to work, the summands need to converge to $0$ in the infinite limit. Of course, showing that in this case is quite easy, and you have already done so.