Check if $\int_1^{\infty\:}\left(e^{-\sqrt{x}}\right)dx$ converges using Convergence Test I could use some help with an homework question:
Using the convergence test, check if the following integral function converges or diverges (no need to calculate the limit itself):
$\int_1^{\infty\:}\left(e^{-\sqrt{x}}\right)dx$
I know that it converges, but i'm requested to find an integral function which upward bounds this function.
 A: Hint: prove that $$e^{-\sqrt{x}} \leq \frac{1}{x^2}$$ whenever $x \gg 1$. You should remark that this follows immediately from a suitable limit at infinity. Then apply the obvious comparison that this inequality suggests.
A: One way to do it would be to turn it into a maclaurin series:
$$e^{-\sqrt{x}} = \sum_{n=0}^\infty \frac{(-\sqrt{x})^n}{n!}$$
$$\implies \int e^{-\sqrt{x}} = \int\sum_{n=0}^\infty \frac{(-\sqrt{x})^n}{n!} = C + \sum_{n=0}^\infty \frac{2x(-\sqrt{x})^n}{(n+2)n!}$$
$$\text{Ratio Test: } \lim_{n \to \infty}\left\lvert \frac{a_{n+1}}{a_n}\right\rvert$$
$$ = \lim_{n \to \infty}\left\lvert \frac{2x(-\sqrt{x})^{n}(-\sqrt{x})}{(n+1)(n+2)n!} *\frac{(n+2)n!}{2x(-\sqrt{x})^n} \right\rvert = \lim_{n \to \infty}\left\lvert \frac{-\sqrt{x}}{n+1} \right\rvert$$
$$\lim_{n \to \infty} \left\lvert \frac{-\sqrt{x}}{n+1} \right\rvert \to 0< 1 \text{ for all values of x } \implies R=\infty, (-\infty,\infty)$$
$$\infty \leq \infty, \space -\infty \leq 0 \leq \infty$$
$$\therefore \int_1^\infty e^{-\sqrt{x}} \text{ converges}$$
