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Convergence of the series : $$ \sum_{n=1}^{\infty} n! \frac{k^n}{n^k}$$

I used ratio test for the case $k \ge 0$ and got that for $k \gt 0$ the series is divergent and for $k=0$ the series is convergent. But I can't figure how to approach the $k\lt 0$ case.

I've thought of Leibnitz test for alternating series but couldn't do anything and I can't prove the convergence or even the divergence of this alternating series. Any help would be appreciated.

Martin Sleziak
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Itachi
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1 Answers1

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For $k=-h<0$ we have

$$\left|n! \frac{k^n}{n^k}\right|= h^nn^hn! \to \infty$$

indeed by root test

$$(h^nn^hn! )^\frac1n = h \cdot (\sqrt[n]{n})^h\cdot \sqrt[n]{n!} \to \infty$$

then the given series doesn't converge for $k<0$ since $a_k \not \to 0$.

user
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  • But you just proved that the given series doesn't converge absolutely...what about it's conditional convergence?? I'm interested in that part...!! – Itachi Sep 26 '20 at 21:23
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    @Pritam We are referring to the necessari condition for convergence that is $a_n \to 0$, this suffice to conclude that the series doesn't converge. – user Sep 26 '20 at 21:26
  • okay, I didn't get that at first that you're trying to show this. So let me just arrange it in a way I can fully understand what your answer is. So you've shown that for the series $\sum (-1)^n a_n$ , $a_n$ doesn't converge to $0$ therefore the n-th term $(-1)^n a_n$ of the given alternating series doesn't tend to $0$ i.e. divergent...am I getting it right? – Itachi Sep 26 '20 at 21:45
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    @Pritam Yes of course, this is a primary requirement for convergence. Then the result is valid not only for the absolute series but also in general. – user Sep 26 '20 at 21:48
  • I understand. I misunderstood what you're trying to show in the answer. Thanks for the help. – Itachi Sep 26 '20 at 21:53
  • Other answers here say the interval of convergence is $(-e,e)$. I tried to compare why the results are different. Maybe I am missing something. – PinkyWay Sep 26 '20 at 22:46
  • @Invisible The series linked is different with $n^n$ at the denominator, here it is $n^k$. – user Sep 26 '20 at 22:48
  • My apologies to you and the OP. – PinkyWay Sep 26 '20 at 22:49