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How should I proceed? What kind of criteria should I apply in this situation?

i tried this thing not really sure if its ok

$\sum_{n=1}^\infty n!^{1/n} > \sum_{n=a}^\infty a^{n^{1/n}} = \sum_{n=a}^\infty a = \infty$ where $a \ge 1$

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  • $\begingroup$ What is a necessary condition for a series to converge? $\endgroup$ – Daniel Fischer Oct 28 '16 at 19:31
  • $\begingroup$ Look for a minor series that diverges. $\endgroup$ – Imago Oct 28 '16 at 19:33
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If $x\geq1$ then $x^{1/n}\geq1$ for every natural $n$. In particular $n!^{1/n}\geq1$. This tells you that the general term doesn't converge to $0$ so the series must diverge.

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