How can we find sum of infinite terms of the series: $$\cot^{-1}(2)+\cot^{-1}\biggl(\frac{9}{2}\biggr)+\cot^{-1}(8)+\cot^{-1}\biggl(\frac{25}{2}\biggr)+\cot^{-1}(18)+ \dots$$
The difference of argument is in A.P. but it does not seem to be valuable. Could someone provide some hint?
Similarly as in Find sum of the Trignomertric series, you can use the identity $$ \DeclareMathOperator{\arccot}{arccot} \arccot(x) - \arccot(y) = \arccot\left(\frac{xy+1}{y-x}\right) $$ to conclude that $$ \arccot \frac{n+2}{n} - \arccot \frac{n}{n-2} = \arccot \frac{n^2}{2} $$ which allows to write your series as a sum of two telescoping series.
HINT:
The series can be written
$$\sum_{n=2}^\infty \text{arccot}\left(\frac{n^2}{2}\right) \tag 1$$
Then, note that
$$\text{arccot}(x)=\frac1x+O\left(\frac{1}{x^3}\right) \tag 2$$
as $x\to \infty$
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
From $(1)$ and $(2)$, we see that the series converges by comparison with $\sum_{n=2}^\infty \frac{1}{n^2}$. Then, we can write the summands as $$\text{arccot}\left(\frac{n^2}{2}\right)=\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)$$which provides a telescoping series. To evaluate the limit, we have $$\begin{align}\sum_{n=2}^N\left(\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)\right)&=\text{arccot}\left(\frac{N+2}{N}\right)\\\\&+\text{arccot}\left(\frac{N+1}{N-1}\right)\\\\&-\text{arccot}\left(3\right)\\\\&\to \frac{\pi}{2}-\text{arccot}(3)\,\,\text{as}\,\,N\to \infty\end{align}$$
