# Question regarding 2 infinite series

First of all, I have to determine whether or not the two infinite series given by $$\displaystyle \sum_{n=4}^{\infty}\left(e^{\frac{\left(-1\right)^{n}}{n}}-1\right)$$ and $$\displaystyle \sum_{n=4}^{\infty}\left(e^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1\right)$$ absolutely converge, conditionally converge or diverge.

Is this a legit statement to contradict absolute convergence of both series?

Using the comparison test with $$\displaystyle \sum_{n=1}^{\infty}n$$, (which diverges) will give us : $$\displaystyle \frac{|e^{\frac{\left(-1\right)^{n}}{n}}-1|}{n}=\left|\frac{e^{\frac{\left(-1\right)^{n}}{n}}}{n}-\frac{1}{n}\right|\underset{n\to\infty}{\longrightarrow}\left|\frac{1}{\infty}-\frac{1}{\infty}\right|=0$$

Therefore, the series $$\displaystyle e^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1$$ does not absolutely converge.

Anyway, I'll be glad to hear some ideas how to prove/disprove conditional convergence of both series. Thanks

$$e^{\frac{(-1)^n}{n}}-1=\frac{(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)$$ The series $$\sum\frac{(-1)^n}{n}$$ converges because it is an alternate series, and $$\sum\mathcal{O}\left(\frac{1}{n^2}\right)$$ converges by comparaison thus $$\sum \left(e^{\frac{(-1)^n}{n}}-1\right)$$ converges. By the same argument, $$e^{\frac{(-1)^n}{\sqrt{n}}}-1=\frac{(-1)^n}{\sqrt{n}}+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$$ The series $$\sum\frac{(-1)^n}{\sqrt{n}}$$ and $$\sum\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$$ converge by the same argument but $$\sum\frac{1}{2n}$$ diverges thus the series $$\sum\left(e^{\frac{(-1)^n}{\sqrt{n}}}-1\right)$$ diverges.

• is this a taylor series ? we have justified in class why it is legit to use it on series/natural numbers, so i cant use it. do you have another idea ? thanks. Apr 17, 2020 at 16:53
• I used the Taylor series $e^x=1+x+\mathcal{O}(x^2)$ and $e^x=1+x+\frac{x^2}{2}+\mathcal{O}(x^3)$ around $x=0$. You can use it with $x=\frac{(-1)^n}{n}$ because $\lim\limits_{n\rightarrow +\infty}\frac{(-1)^n}{n}=0$, same with $\frac{(-1)^n}{\sqrt{n}}$. Apr 17, 2020 at 16:56
• can you explain why we know that $\sum\mathcal{O}\left(\frac{1}{n^{2}}\right)$ converges? i mean i know that $\sum\left(\frac{1}{n^{2}}\right)$ converges, but how can i prove in a formal way that $\sum\left(\frac{1}{n^{2}}\right)$ converges too? im sorry im not familier with proving convergence with taylor series Apr 17, 2020 at 18:56
• It is a comparaison : if $u_n=\mathcal{O}(1/n^2)$ there exists $C>0$ such that $u_n\leqslant\frac{C}{n^2}$ and since $\sum\frac{C}{n^2}<+\infty$ you have $\sum|u_n|<+\infty$. Apr 17, 2020 at 19:02
• why the reminder in the second series is $O\left(\frac{1}{n^{\frac{3}{2}}}\right)$ and not $O\left(\frac{\left(-1\right)^{n}}{n^{\frac{3}{2}}}\right)$ ? Apr 17, 2020 at 19:20

\begin{aligned} \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1=\frac{\left(-1\right)^{n}}{n}\int_{0}^{1}{\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}x}\,\mathrm{d}x}&=\frac{\left(-1\right)^{n}}{n}+\frac{\left(-1\right)^{n}}{n}\int_{0}^{1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}x}-1\right)\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{n}+\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\\ \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1 &=\frac{\left(-1\right)^{n}}{n}+v_{n} \end{aligned}

With $$v_{n}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) \cdot$$

From that remains the following : \begin{aligned} \left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right|&\geq\left|\left|\frac{\left(-1\right)^{n}}{n}\right|-\left|v_{n}\right|\right|=\frac{1}{n}-\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\end{aligned}

Since $$\sum\limits_{n\geq 1}{\frac{1}{n}}$$ diverges, $$\sum\limits_{n\geq 1}{\left(\frac{1}{n}-\frac{1}{n^{2}}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}xy}\,\mathrm{d}y}\,\mathrm{d}x}\right)}$$ will diverge, and thus, by comparaison, $$\sum\limits_{n\geq 1}{\left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right|}$$ diverges.

Now since $$\sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{n}}$$ converges by the AST, and $$\sum\limits_{n\geq 1}{v_{n}}$$ converges by comparaison test, we have that $$\sum\limits_{n\geq 1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1\right)}$$ converges.

Now, we can do pretty much the same thing for the second one :

\begin{aligned} \mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1=\frac{\left(-1\right)^{n}}{\sqrt{n}}\int_{0}^{1}{\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}x}\,\mathrm{d}x}&=\frac{\left(-1\right)^{\sqrt{n}}}{n}+\frac{\left(-1\right)^{n}}{\sqrt{n}}\int_{0}^{1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}x}-1\right)\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{n}\int_{0}^{1}{\int_{0}^{1}{x\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xy}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+\frac{1}{n}\int_{0}^{1}{\int_{0}^{1}{x\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xy}-1\right)\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+\frac{\left(-1\right)^{n}}{n\sqrt{n}}\int_{0}^{1}{\int_{0}^{1}{\int_{0}^{1}{x^{2}y\,\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}xyt}\mathrm{d}t}\,\mathrm{d}y}\,\mathrm{d}x}\\ \mathrm{e}^{\frac{\left(-1\right)^{n}}{n}}-1 &=\frac{\left(-1\right)^{n}}{\sqrt{n}}+\frac{1}{2n}+w_{n} \end{aligned}

With $$w_{n}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n\sqrt{n}}\right) \cdot$$

We have that $$\sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n}}{\sqrt{n}}}$$ converges by the AST, and $$\sum\limits_{n\geq 1}{w_{n}}$$ converges by comparaison test, but with $$\sum\limits_{n\geq 1}{\frac{1}{n}}$$ being divergent. We have that $$\sum\limits_{n\geq 1}{\left(\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1\right)}$$ diverges.

And thus, $$\sum\limits_{n\geq 1}{\left|\mathrm{e}^{\frac{\left(-1\right)^{n}}{\sqrt{n}}}-1\right|}$$ also diverges.

• how cone $w_{n}=O\left(\frac{1}{n\sqrt{n}}\right)$ and not $w_{n}=O\left(\frac{\left(-1\right)^{3n}}{n\sqrt{n}}\right)$ ? Apr 17, 2020 at 19:18
• @Waizman They are the same thing, since $\left(-1\right)^{n}$ is bounded. Multiplying by a bounded function won't change anything, asymptotically speaking. Apr 17, 2020 at 19:23