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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

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2 Answers 2

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$$ e^{\frac{(-1)^n}{n}}-1=\frac{(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^2}\right) $$ 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.

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  • $\begingroup$ 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. $\endgroup$
    – FreeZe
    Apr 17, 2020 at 16:53
  • $\begingroup$ 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 $\frac{(-1)^n}{\sqrt{n}}$. $\endgroup$
    – Tuvasbien
    Apr 17, 2020 at 16:56
  • $\begingroup$ 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 $\endgroup$
    – FreeZe
    Apr 17, 2020 at 18:56
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    $\begingroup$ 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 $\sum|u_n|<+\infty$. $\endgroup$
    – Tuvasbien
    Apr 17, 2020 at 19:02
  • $\begingroup$ 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) $ ? $\endgroup$
    – FreeZe
    Apr 17, 2020 at 19:20
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\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.

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  • $\begingroup$ 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) $ ? $\endgroup$
    – FreeZe
    Apr 17, 2020 at 19:18
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    $\begingroup$ @Waizman They are the same thing, since $ \left(-1\right)^{n} $ is bounded. Multiplying by a bounded function won't change anything, asymptotically speaking. $\endgroup$
    – CHAMSI
    Apr 17, 2020 at 19:23

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