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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?

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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
up vote 7 down vote accepted

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.

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I think the Alternate series test is easier for $k=-1$ ( ). – Listing May 6 '11 at 6:51

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.

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@Luke: I note that you should consider the endpoints, when k = 1 and -1, separately. – mixedmath May 6 '11 at 5:50

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