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

Hint:
Near $+\infty$, $\;n^2=o\bigl(\mathrm e^{\sqrt n}\bigr)$, whence $\;\mathrm e^{-\sqrt n}=o\Bigl(\dfrac1{n^2}\Bigr)$.
