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Determine if the series $\sum \alpha_n$ 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.

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    $\begingroup$ Since $e^{-u}=\frac 1{e^u}$ your first inequality will suffice. $\endgroup$ – abiessu Oct 13 '18 at 15:55
  • $\begingroup$ 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? $\endgroup$ – Nikolaos Skout Oct 13 '18 at 16:14
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    $\begingroup$ 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 $n\to \infty$. $\endgroup$ – Mark Viola Oct 13 '18 at 17:39
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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.

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    $\begingroup$ Well done. (+1) $\endgroup$ – Mark Viola Oct 13 '18 at 17:45
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$$\alpha_n =\frac{1}{n}\int_{1}^{+\infty} \frac{z^{1/n}}{z} e^{-z}=\frac{1}{en}\int_{0}^{+\infty}(z+1)^{1/n}\frac{dz}{e^z(z+1)} $$ 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.

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