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
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.
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