Determine if the series $\sum \alpha_n$ converge, where:


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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 2
    $\begingroup$ Well done. (+1) $\endgroup$ – Mark Viola Oct 13 '18 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.