I want to make sure I am correctly understanding the intuition behind the comparison theorem. Suppose I'm given the following integral:
$$\int_0^{\infty}\frac{\arctan{x}}{2+e^x}$$ First, I notice that $\arctan{x}$ lies between $0$ and $\frac{\pi}{2}$ for $0\leq x<\infty$. From this I have that $\arctan{x}<\frac{\pi}{2}$, such that $$\frac{\arctan{x}}{2+e^x}<\frac{\pi}{2}\bigg(\frac{1}{2+e^x}\bigg)=\frac{\pi}{4+2e^x}.$$ I can make this expression on the right even bigger by making the denominator smaller. That is, we have the following inequality:
$$\frac{\arctan{x}}{2+e^x}<\frac{\pi}{4+2e^x}<\frac{\pi}{2e^x}.$$ It remains to investigate the convergence of $$\frac{\pi}{2e^x}.$$ We have that $$\int_0^{\infty}\frac{\pi}{2e^x}=\frac{\pi}{2}\int_0^{\infty}e^{-x}=\frac{\pi}{2}\bigg(\lim_{t\to\infty}-e^{-x}\bigg|_0^t\bigg)=\frac{\pi}{2}\bigg(\lim_{t\to\infty}-e^{-t}+1\bigg).$$ The limit is convergent, and because our upper bound converges, so too must our original function by the comparison test. Does this seem like a good way of going about solving such problems?