Sum of infinite terms of $\cot^{-1}(2)+\cot^{-1}\bigl(\frac{9}{2}\bigr)+\cot^{-1}(8)+\cot^{-1}\bigl(\frac{25}{2}\bigr)+\cot^{-1}(18)+...$ 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?
 A: 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.
A: 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 $$\begin{align}\sum_{n=2}^N\left(\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)\right)&=\text{arccot}\left(\frac{N+2}{N}\right)\\\\&+\text{arccot}\left(\frac{N+1}{N-1}\right)\\\\&-\text{arccot}\left(3\right)\\\\&\to \frac{\pi}{2}-\text{arccot}(3)\,\,\text{as}\,\,N\to \infty\end{align}$$ 

