Convergence of a series with general term $u_n=\int_0^{\infty}e^{-x^n}dx$ I  would like to find if the series $\displaystyle \sum_{n=1}^{\infty}u_n$ is convergent or divergent where $$u_n=\int_0^{\infty}e^{-x^n}dx. $$
I've tried to obtain $v_n$ with explicit form such that $u_n\leq v_n$ and $\displaystyle \sum_{n=1}^{\infty}v_n$ converges, but I didn't find it. 
 A: You are looking for a majorant $v_n$. Let's instead, as Kainui's comment (and picture) suggest, try to do the opposite: $$u_n\ge \int_0^1\exp(-x^n) \ d x \ge \int_0^1 e^{-x}\ dx. $$ This follows since when $0<x<1$ we have $x^n< x$ and $t\mapsto e^{-t}$ is a decreasing function. Thus $$ u_n\ge 1/e $$ so what can you conclude?
A: For the divergence test, just take the limit as n approaches infinity we get:
$ \LARGE lim_{n \rightarrow \infty} \int_0^\infty  e^{-x^n}dx  \\ \LARGE lim_{n \rightarrow \infty} \int_0^1  e^{-x^n}dx +lim_{n \rightarrow \infty} \int_1^\infty  e^{-x^n}dx \\ \LARGE \int_0^1 1 dx + \int_1^\infty0dx =1  $
Since for x less than 1,  $x^\infty$ approaches 0 giving us $e^0=1$ and for x greater than 1, we have $e^{-\infty}$ which approaches 0.
So it fails the divergence test since the limit is 1, and it has to be 0 to converge.
A: Note that, for $x \geq 0, n \in \mathbb{N}$, you have
$$ x \leq (\log 2)^{\frac{1}{n}} \Leftrightarrow e^{-x^n} \geq \frac{1}{2}$$
Hence
$$\int_0^{+\infty} e^{-x^n} dx \geq \int_0^{(\log 2)^{\frac{1}{n}}} e^{-x^n} dx \geq 
\int_0^{(\log 2)^{\frac{1}{n}}} \frac{1}{2} dx = \frac{1}{2} (\log 2)^{\frac{1}{n}} \to \frac{1}{2}
$$
as $n \to + \infty$. So your series is divergent.
