Study the convergence of the Series $\sum_{n=0}^{\infty} e^{-\sqrt{n}}$ Study the convergence of the Series $\sum_{n=0}^{\infty} e^{-\sqrt{n}}$ The only thing I know is that $e^{-\sqrt{x}}$ is strictly decreasing. I also know that the only method here to use is the Comparison or limit criteria, but i don't know to what sequence compare it to. Thanks in advance
 A: Hint:
$$e^{-\sqrt{n}}\leq \frac{1}{n^2} \quad \text{for sufficiently large }n$$
Showing that is easy if you use Taylor expansion of $e^x$.
A: @mertunsal's solution is fine, but I will complement it with the best tips I can give you about checking the convergence/divergence of series.
First, be sure that you know the convergence/divergence of all basic series (for example, the behavior of the geometric series and that $\sum \frac{1}{j^\alpha}$ converges iff $\alpha > 1$).
Secondly be familiar with the growth speed of all basic functions, it must be automatic for you that $\log_a(n)$ < polynomials of grade 1 < polynomials of grade 2 < polynomials of grade 3 < polynomials of grade 4 < ... < $a^n < n! < n^n$ for $n$ big enough (obs: I am assuming $a>1$).
The critical point for understanding the convergence/divergence of an expression like this one is to feel how it does behave compared to those basics above, so you can use theorems of comparison.
The function $\exp(\sqrt(n))$ is not in the list of the basic ones, but as we are aware that $\exp(n) > p(n)$ for any p polynomial (with a positive leading coefficient), then $\exp(\sqrt{n}) > p(\sqrt{n})$. So, considering $p(n)=n^4$ you will get $\exp(\sqrt{n}) > n^2$ for $n$ big enough. Because $\sum \frac{1}{n^2}$ converges, then it is easy to conclude the convergence of the your series.
A final tip: sometimes, you might need to justify with more details the inequality (either because you are a student and your professor is a pain in the ass - or because your expression is complex and you just have a feeling of how it grows instead of a rigorous argument). In these cases, often to check the limit of the ratio between the two expressions solves the issue. For example, in your case, you would need to check that $\exp(\sqrt{n}) > n^2$. Then you could simply show that $\lim_{n\to\infty} \frac{\exp(\sqrt{n})}{n^2} = \infty$ (just use L'Hospital or another standard method), so you can conclude that $\exp(\sqrt{n}) > n^2$ for $n$ big enough. This method is very powerful and useful at hard issues.
Well, I hope it helps. And apologize me for this cold murder of the English Language, I am not a native speaker and I wrote it in a hurry.
A: If you look at $\int_{0}^{\infty}e^{-\sqrt{x}}dx$, set $\sqrt{x}=t$, you get $\frac{1}{2}\int_{0}^{\infty}t e^{-t}dt$, which is half the expectation of exponential random variable with parameter $\lambda=1$, i.e. $\frac{1}{2}$, so your sum converges.
