Sum of infinite terms of $\cot^{-1}(2)+\cot^{-1}\bigl(\frac{9}{2}\bigr)+\cot^{-1}(8)+\cot^{-1}\bigl(\frac{25}{2}\bigr)+\cot^{-1}(18)+...$

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 $$\begin{align}\sum_{n=2}^N\left(\text{arccot}\left(\frac{n+2}{n}\right)-\text{arccot}\left(\frac{n}{n-2}\right)\right)&=\text{arccot}\left(\frac{N+2}{N}\right)\\\\&+\text{arccot}\left(\frac{N+1}{N-1}\right)\\\\&-\text{arccot}\left(3\right)\\\\&\to \frac{\pi}{2}-\text{arccot}(3)\,\,\text{as}\,\,N\to \infty\end{align}$$