Convergence of series $\sum\limits_{n=1}^\infty\int\limits_{1}^{+\infty}e^{-x^n}\,dx$ 
Determine if the series $\sum\limits_{n=1}^\infty \alpha_n$ converge, where
$$\alpha_n=\int\limits_{1}^{+\infty}e^{-x^n}\,dx.$$

Attempt. Ι am pretty sure the inequalities $$e^{-x^n}\leq \frac{1}{1+x^n}$$
and $e^{-x^n}\geq 1-x^n$ will be useful (the first one I believe more, which gives convergence for the series though, which I am not sure if it is correct).
Thanks for the help.
 A: $$\alpha_n =\frac{1}{n}\int_{1}^{+\infty} \frac{z^{1/n}}{z} e^{-z}=\frac{1}{en}\int_{0}^{+\infty}(z+1)^{1/n}\frac{dz}{e^z(z+1)} $$
is not a summable term since the dominated convergence theorem ensures 
$$ \lim_{n\to +\infty}\int_{0}^{+\infty}(z+1)^{1/n}\frac{dz}{e^z(z+1)}=\int_{0}^{+\infty}\frac{dz}{e^z(z+1)}\approx\frac{31}{52} $$
and the harmonic series is divergent.
A: Note that
$$\alpha_n=\int_1^\infty\,\exp\left(-x^n\right)\,\text{d}x=\frac{1}{n}\,\int_1^\infty\,t^{-\left(1-\frac{1}{n}\right)}\,\exp(-t)\,\text{d}t\geq \frac{1}{n}\,\int_1^\infty\,\frac{\exp(-t)}{t}\,\text{d}t\,,$$
by setting $t:=x^{\frac1n}$.  Therefore,
$$\alpha_n\geq \frac{\lambda}{n}\,,\text{ where }\lambda:=\int_1^\infty\,\frac{\exp(-t)}{t}\,\text{d}t=-\text{Ei}(-1)\approx 0.21938\,.$$
Here, $\text{Ei}$ is the exponential integral.  (We do not need the value of $\lambda$, just that it is a finite positive real number.)  Thus, the sum $\sum\limits_{n=1}^\infty\,\alpha_n$ diverges due to divergence of the harmonic series.

On the other hand, we can also see that
$$\alpha_n\leq \frac{1}{n}\,\int_1^\infty\,\exp(-t)\,\text{d}t=\frac{1}{n}\,\exp(-1)=\frac{1}{n\,\text{e}}\,.$$
Therefore, $\alpha_n \in \Theta\left(\dfrac{1}{n}\right)$ as $n\to\infty$, with
$$-\text{Ei}(-1)\leq \liminf_{n\to\infty}\,n\,\alpha_n\leq \limsup_{n\to\infty}\,n\,\alpha_n\leq \frac{1}{\text{e}}\,.$$
I expect that $\lim\limits_{n\to\infty}\,n\,\alpha_n$ exists, though, and conjecture that the limit is precisely $-\text{Ei}(-1)$.   

Let $f:[1,\infty)\to\mathbb{R}$ and, for each $n\in\mathbb{Z}_{>0}$, $f_n:[1,\infty)\to\mathbb{R}$ be the functions defined by $$f(t):=\frac{\exp(-t)}{t}\text{ and }f_n(t):=t^{-\left(1-\frac{1}{n}\right)}\,\exp(-t)$$ for all $t\geq 1$.  Then, $f_n\to f$ as $n\to \infty$ pointwise, $\left|f_n\right|=f_n\leq g$, where $g:[1,\infty)\to\mathbb{R}$ is an integrable function given by $$g(x)=\exp(-t)\text{ for all }t\geq 1\,,$$ and $$\begin{align}\int_1^\infty\,\left|f_n(t)-f(t)\right|\,\text{d}t&=\int_1^\infty\,\left(t^{\frac{1}{n}}-1\right)\,\frac{\exp(-t)}{t}\,\text{d}t\\&\leq \int_1^\infty\,\left(t^{\frac{1}{n}}-1\right)\,\exp(-t)\,\text{d}t\\&\leq\Gamma\left(1+\frac{1}{n}\right)-\Gamma(1)\underset{n\to\infty}{\longrightarrow}0\,,\end{align}$$ where $\Gamma$ is the usual gamma function (which is continuous).  By the Dominated Convergence Theorem, $$\lim_{n\to\infty}\,\int_1^\infty\,f_n(t)\,\text{d}t=\int_1^\infty\,f(t)\,\text{d}t\,.$$ Therefore, $n\,\alpha_n$ does indeed converge to $-\text{Ei}(-1)$, as $n$ grows to infinity.

