Does $\sum_{n=1}^{\infty} \frac{(-1)^n}{\arctan(n)}$ converge?

Question

I can't use Leibniz since $\lim_{n\to\infty} \frac{1}{\arctan(n)} = \frac 2 \pi \ne 0$ (and it seems to diverge anyway, but how can I prove it?)

Answer 1

Let $a_{n}=(-1)^{n}/\tan^{-1}n$, if it were convergent, then $\lim_{n\rightarrow\infty}a_{n}=0$, and hence $\lim_{n\rightarrow\infty}|a_{n}|=0$, but you have disproved the last one.

Answer 2

With $$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{\arctan n}=\sum_{n\text{ is odd}}^{\infty}\left(\dfrac{1}{\arctan n+1}-\dfrac{1}{\arctan n}\right)$$ and using the fact $\arctan n\geq\dfrac{\pi n}{4}$, then $$\dfrac{1}{\arctan n+1}-\dfrac{1}{\arctan n}=\dfrac{\arctan\dfrac{1}{n^2+n-1}}{\arctan (n+1)\arctan n}\leq\dfrac{\dfrac{\pi}{4}}{\dfrac{\pi}{4}(n+1)\dfrac{\pi}{4}n}\leq\dfrac{4}{\pi}\dfrac{1}{n^2}$$ so the sum will be less than $\dfrac{\pi}{2}$.