# Convergence of $\sum\limits_{n=1}^{\infty} k^{1/n}$

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.

• Does $k^{1/n}\to0$ as $n\to\infty$? – Lord Shark the Unknown Apr 25 '17 at 5:37
• It does not seem to. I think the terms converge to 1, but I ain't sure about their sum. – 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 $\displaystyle \lim_{n \rightarrow \infty} k^{1/n}$?
• 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 – Red Flag Apr 25 '17 at 5:50
• Yep, the value for $k$ is pretty irrelevant, except if we had $k = 0$. – Kaj Hansen Apr 25 '17 at 5:50
• Edited my previous comment to exclude $k=0$ :) – Red Flag Apr 25 '17 at 5:52