$\lim_{n\rightarrow\infty} \sum_{k=1}^n \text{arccot}(2k^2)$

Question

I think this can be solved using the sandwiching theorem but I have not been able to find appropriate series to sandwich with. The trivial substitution yields that the limit is between $0$ and $\infty$ which is not very helpful.

Answer 1

Note $\cot^{-1}(2k^2)=\arctan(\frac{1}{2k^2})$, so we wish to find $\sum_{k \ge 1}\arctan(\frac{1}{2k^2})$. This is $\sum_{k \ge 1}\arctan(\frac{2}{4k^2})$. This is $\sum_{k \ge 1}\arctan(\frac{1}{2k-1})-\sum_{k \ge 1}\arctan(\frac{1}{2k+1})$. Note the comment of J.G. above. Hence the answer is $\arctan 1$, that is $\frac{\pi}{4}$. I reference this website that helped me:

https://www.physicsforums.com/threads/find-the-sum-of-arctan-1-2n-2.80196/#google_vignette

Cheers.