Determine if $\int_1^\infty \arctan(e^{-x})dx$ converges or diverges I want to check if the sequence $\sum_{n=1}^\infty [\sin(n) \cdot \arctan(e^{-n})]$ converges absolutely, by condition, or not.
We know that $\sum_{n=1}^\infty [\sin(n) \cdot \arctan(e^{-n})] \le \sum_{n=1}^\infty \arctan(e^{-n})$, so I want to check with the integral test.
Now I need to determine if $\int_1^\infty \arctan(e^{-x})dx$ converges or diverges, but I don't know how to do it.
Will appreciate your help with that (even if there is a better solution for the sequence, I'd love to know about the integral).
Thanks!
 A: If we can prove that $\arctan e^{-x}<e^{-x}$ then the integral converges by the comparison test since $\int_1^\infty e^{-x}=1/e<\infty$.
Proving this inequality with the mean value theorem as Severin Schraven in the comments suggested: For some $c\in(0,x)$
$$\frac{\arctan(x)-\arctan(0)}{x-0}=\frac1{1+c^2}\Rightarrow\arctan x \leqslant x.$$ Or maybe even faster, observe that $\arctan x=\int_0^x\frac{1}{u^2+1}\,\mathrm{d}u$ and that $x=\int_0^x 1\mathrm{d}u$. Now deduce the inequality easily from the comparison test.
A: To finish your problem, it is enough, as you pointed out, to check for convergence of the integral $\int^\infty_0\arctan(e^{-x})\,dx$.
The change of  variables $u=e^{-x}$ gives
$$\int^A_1\arctan(e^{-x})\,dx=\int^A_1\arctan(e^{-x})\frac{e^{-x}}{e^{-x}}\,dx=\int^{e^{-1}}_{e^{-A}}\frac{\arctan u}{u}\,du$$
The function $\phi:u\mapsto\frac{\arctan u}{u}$ can be extended continuously at $u=0$ by setting $\phi(0)=1$ for $\lim_{u\rightarrow0}\phi(u)=1$ (L'Hospital rule for example).
If you are considering improper Riemann integrals integrability, integrability follows from letting $A\rightarrow\infty$ since $\phi$ is Riemann integrable in $[0,1]$; in the sense of  Lebesgue,  integrability follows from monotone convergence ( $\arctan$ and $\phi$ are nonnegative on $[0,\infty)$). In either case,
$$\int^\infty_1\arctan(e^{-x})\,dx=\int^{e^{-1}}_0\frac{\arctan u}{u}\,du<\infty$$
