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I was wondering why in some exercise I found it says $\sum_{n=0}^{\infty}(-n)^3e^{{-n}^{-4}}$ absolutely converges .

If it is absolutely convergent then it means that $\sum_{n=0}^{\infty}(n)^3e^{{-n}^{-4}}$ converges . I tried the ratio test for series but it becomes harsh. Also I tried to develop a pattern by developing the sequence of partial sums but it seem the series is growing a lot. I don't find also a bound for each term of the sequence of partial sum in order to apply the Weierstrass M test. I need help in order to figure out and prove the series converges. Thank you so much.

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    $\begingroup$ What's up with $e^{{-n}^-4}$? $\endgroup$ – zhw. Nov 22 '15 at 20:02
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    $\begingroup$ Unless Cervus accepts the answer, the system will keep returning this to the "top questions" page. $\endgroup$ – GEdgar Mar 19 at 12:20
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The sum is $$ \sum^{\infty}_{n=0} (-n)^3 \cdot \mathrm{exp}(-\frac{1}{n^4}) = \sum^{\infty}_{n=0} (-1)^3(n)^3 \cdot \mathrm{exp}(-\frac{1}{n^4}) = -\sum^{\infty}_{n=0} (n)^3 \cdot \mathrm{exp}(-\frac{1}{n^4})$$ Since the terms $$ n^3 \cdot \mathrm{exp}(-\frac{1}{n^4}) \rightarrow \infty $$  when $n \rightarrow \infty$ the sum isn't aboslute convergent, nor convergent.

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  • $\begingroup$ Nor even convergent right? @fritzenbauer $\endgroup$ – Cervus Nov 22 '15 at 20:01
  • $\begingroup$ No you're right, it isn't converging at all! $\endgroup$ – fritzenbauer Nov 22 '15 at 20:03
  • $\begingroup$ Thanks I appreciate it. @fritzenbauer $\endgroup$ – Cervus Nov 22 '15 at 20:06
  • $\begingroup$ Are you sure that it said $e^{-n^{-4}}$ and not $e^{-n^4}$ ? $\endgroup$ – DanielWainfleet Nov 23 '15 at 1:07
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The sequence is not absolutely convergent since$$\left|(-n)^3e^{-n^{-4}}\right|=n^3e^{-n^{-4}}\sim n^3\quad,\quad n\to \infty$$

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