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I have a few questions about applying Leibniz (Alternating Series) Test and hope you can help me with that. As an example I will use the following series : $$\sum_{i=1}^n ((-1)^{9n}n^2e^{\frac{-n^2}{3}})$$

  • When looking at $\lim_{n\to \infty}a_n=\lim_{n\to \infty}(-1)^{9n}n^2e^{\frac{-n^2}{3}}$ I don't quite understand what can be done with $(-1)^{9n}$. Without it , I could have easily found the limit using L'Hospital's rule and proven that it is indeed approaches $0$ (as needed). Should I instead look at $\lim_{n\to \infty}|a_n|$ then?
  • I also don't understand how I can prove that $a_n>0$ if it actually depends on $(-1)^{9n}$. Am I doing something wrong here as well?

I don't need the complete solution as know that it is converges. I do hope, however , that you can help me to understand this test better.

P.S. (I remember about the decreasing series condition, but I decided to skip it here as it is not related to my main questions)

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3 Answers 3

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Well, $(-1)^{9n}=(-1)^n$. Besides, $\left(n^2e^{-\frac{n^2}3}\right)_{n\in\mathbb N}$ is decreasing. So…

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$$ (-1)^{9n} = (-1)^n(-1)^{8n} = (-1)^n \text{.} $$

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  • $\begingroup$ I do understand that, but how can it help me prove that $\lim_{n\to \infty}a_n$ approaches 0? My main question is basically about the sign there. How can I find the limit if the sign depends on n which approaches infinity? $\endgroup$
    – Alex.Kh
    Commented Mar 20, 2020 at 21:38
  • $\begingroup$ @Alex.Kh Well you said " I don't quite understand what can be done with $(-1)^{9n}$. Without it , I could have easily found the limit using L'Hospital's rule and proven that it is indeed approaches $0$ (as needed)." so hat directly answers the first point you've raised, right? $\endgroup$ Commented Mar 20, 2020 at 22:05
  • $\begingroup$ @Alex.Kh : You say you do not need any help showing the sequence decreases. "(I remember about the decreasing series condition, but I decided to skip it here as it is not related to my main questions)". This is the alternating series test. There should be a component that causes the sign to alternate and the remainder which should (eventually) go to zero. You claim to understand the second part. $\endgroup$ Commented Mar 20, 2020 at 22:40
  • $\begingroup$ @EricTowers are you saying than that I just have to forget for a moment about the alternating part and simply disregard the sign during analysis ? If so, it is not hard to understand the remaining solution. $\endgroup$
    – Alex.Kh
    Commented Mar 20, 2020 at 22:47
  • $\begingroup$ @Alex.Kh : Do you not have a clear statement of the alternating series test? $\endgroup$ Commented Mar 20, 2020 at 23:00
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One more:

$(-1)^{9n}=((-1)^9)^n=(-1)^n$.

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