2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ What about $k=-1$? $\endgroup$ – Shai Covo May 6 '11 at 5:34
  • 1
    $\begingroup$ 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 $k \neq 1$ $\endgroup$ – user17762 May 6 '11 at 6:04
  • $\begingroup$ Look for "radius of convergence" in your calculus textbook. $\endgroup$ – GEdgar May 6 '11 at 13:30
7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

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

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.