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.

  • 1
    $\begingroup$ 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

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 $\displaystyle \lim_{n \rightarrow \infty} k^{1/n}$?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.