# For what real values of $k$ does the series $\sum\frac{k^n}{\sqrt{n}}$ converge?

I think it's $|k| < 1$, but I don't know how to prove it. It's either that or it never converges. $\sum\frac{1}{\sqrt{n}}$ obviously diverges, but can't an exponential beat it and make the sum finite?

• What about $k=-1$? – Shai Covo May 6 '11 at 5:34
• Just as a side note, it can be proved with a little more effort (using Cauchy root test and generalized alternate test) that if $k \in \mathbb{C}$ the series converges for all $|k| \leq 1$ and $k \neq 1$ – user17762 May 6 '11 at 6:04
• Look for "radius of convergence" in your calculus textbook. – GEdgar May 6 '11 at 13:30

If you do D'alembert's test you will have $$\lim_{n \to \infty}{\left|\frac{\frac{k^{n+1}}{\sqrt{n+1}}}{\frac{k^n}{\sqrt{n}}}\right|} = \lim_{n \to \infty}{\left|\frac{k^{n+1}\sqrt{n}}{k^n\sqrt{n+1}}\right|} = \lim_{n \to \infty}{|k|\sqrt{\frac{n}{n+1}}} = |k|$$ So this tells you that the series converges for $|k| < 1$ and diverges for $|k| > 1$, for $k = 1$ it diverges like you said, and for $k = -1$ it is easy to see using Dirichlet's test that the series converges.
You are correct. I would recommend using a comparison test to the geometric series $\sum k^n$ to show convergence and comparison to the series $\sum \frac{1}{\sqrt{n}}$ to show divergence.