Convergence of $\sum_{n=1}^{\infty}e^{-\sqrt{n}}$ using the integral test Given the series : $$\sum_{n=1}^{\infty}e^{-\sqrt{n}}$$ Determine if convergent or divergent.
The function is positive and monotonically decreasing function so I've used the "Integral Test" 
$$\int_{1}^{\infty}e^{-\sqrt{x}}$$
then:$$ e^{-\sqrt{x}}\leq e^{\sqrt{x}}\leq e^{x}$$
$$\int_{1}^{\infty}e^{x}dx$$ which is clearly Divergent but the answer for some reason is convergent 
Edit: Sorry for the misinterpretation. I forgot that if $f(x)>g(x)$ and $f(x)$ is divergent, it doesn't necessarily mean that $g(x)$ is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this

I usually think that Integral Test is the least elegant way to show the convergence of a sequence. I do have to admit that it is convenient, simple and powerful, though.
In order to show convergence of series involving negative power of $e$, using the Taylor Expansion of $e^x$ is a good way.
Notice that
$$e^\sqrt{n} = \sum_{k=0}^\infty \frac{(\sqrt{n})^k}{k!} > \frac{n^2}{4!}$$
we have
$$ \sum_{n=1}^\infty e^{-\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{e^\sqrt{n}} < \sum_{n=1}^\infty \frac{4!}{n^2} $$
and then you should be able to prove that the series converges.
