Is this :$\sum_{n=1}^{\infty } \ \frac{\tan(\frac{1}{n!})}{\arctan (n!)}$ convergent sum?

Question

How do I evaluate this sum :$$\displaystyle\sum_{n=1}^{\infty } \ \frac{\tan(\frac{1}{n!})}{\arctan ({n!})}$$ if it is convergent ?.

Note: I think the limit of it's general term is $0$ as shown here in WA. and i will surprised if it is convergent

Note: I edited the question beacuse i meant $ arctan(n!)$ in the denominator

Answer 1

One has, as $n \to \infty$, $$ \frac{\tan \frac1{n!}}{\arctan (n!)}=\frac{\tan \frac1{n!}}{\frac{\pi}2-\arctan \frac1{n!}}\sim \frac2{\pi} \cdot \frac1{n!} $$ then by the comparison test the series $ \displaystyle \sum_{n\ge1}\frac{\tan \frac1{n!}}{\arctan n!}$ is convergent.

Answer 2

$\tan\dfrac1{n!}\sim_\infty\dfrac1{n!}$, $\; \arctan(n!)\sim_\infty\dfrac\pi2$, hence $$u_n\sim_\infty\frac 2{\pi n!},$$ which converges (to $2\mathrm e/\pi$).