# 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.

• Since $e^{-u}=\frac 1{e^u}$ your first inequality will suffice. Commented Oct 13, 2018 at 15:55
• Thank you. Since we go for convergence with comparison test, as for $\int_{1}^{+\infty}\frac{dx}{1+x^n}$, inequality $1+x^n\geq x^n$ does not work since it leads to $\sum \frac{1}{n}.$ What other inequalites could I use? Commented Oct 13, 2018 at 16:14
• The bound $e^{-x^n}\le \frac{1}{1+x^n}$ is not useful here inasmuch as $$\int_0^\infty \frac{1}{1+x^n}\,dx=\frac{\pi}{n\sin(\pi/n)}$$which approaches $1$ as $n\to \infty$. Commented Oct 13, 2018 at 17:39

## 2 Answers

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.

• Well done. (+1) Commented Oct 13, 2018 at 17:45

$$\alpha_n =\frac{1}{n}\int_{1}^{+\infty} \frac{z^{1/n}}{z} e^{-z}=\frac{1}{en}\int_{0}^{+\infty}(z+1)^{1/n}\frac{dz}{e^z(z+1)}$$ 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.