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?
For what real values of $k$ does the series $\sum\frac{k^n}{\sqrt{n}}$ converge?
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calculus
real-analysis
sequences-and-series
convergence-divergence
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0Look for "radius of convergence" in your calculus textbook. – 2011-05-06
2 Answers
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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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1I think the Alternate series test is easier for $k=-1$ ( http://en.wikipedia.org/wiki/Leibniz_test ). – 2011-05-06
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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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0@Luke: I note that you should consider the endpoints, when k = 1 and -1, separately. – 2011-05-06