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Does $\sum\limits_{n=1}^{\infty} k^{1/n}$ converge when $k<1$ ??? How to show whether it does or does not then? Integral test or comparison test with $k^n$ does not seem to work.

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    $\begingroup$ Does $k^{1/n}\to0$ as $n\to\infty$? $\endgroup$ – Lord Shark the Unknown Apr 25 '17 at 5:37
  • $\begingroup$ It does not seem to. I think the terms converge to 1, but I ain't sure about their sum. $\endgroup$ – Red Flag Apr 25 '17 at 5:38
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If $\displaystyle \sum a_n$ converges, it is necessary that $a_n \rightarrow 0$. What is $\displaystyle \lim_{n \rightarrow \infty} 1/n$? Given that $k^x$ is continuous, what is $\displaystyle \lim_{n \rightarrow \infty} k^{1/n}$?

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  • $\begingroup$ It is 1. So the sum diverges? $\endgroup$ – Red Flag Apr 25 '17 at 5:41
  • $\begingroup$ Indeed @RedFlag, see here: en.wikipedia.org/wiki/Term_test $\endgroup$ – Kaj Hansen Apr 25 '17 at 5:42
  • $\begingroup$ Also a good thing I can note is that this diverges not only for $k<1$, but for all real values of k except 0 $\endgroup$ – Red Flag Apr 25 '17 at 5:50
  • $\begingroup$ Yep, the value for $k$ is pretty irrelevant, except if we had $k = 0$. $\endgroup$ – Kaj Hansen Apr 25 '17 at 5:50
  • $\begingroup$ Edited my previous comment to exclude $k=0$ :) $\endgroup$ – Red Flag Apr 25 '17 at 5:52

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