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 May 14th, 2009, 06:39 AM #1 Newbie   Joined: Jun 2008 Posts: 20 Thanks: 0 Alternating Series Hey, I have a question for you guys. $\displaystyle\sum (-1)^{n+1 }\ \frac{ln n}{n\sqrt{n}}$ It's obviously convergent due to the AST, but I can't determine whether it's conditionally so or absolutely so. I can't use the integral test, because substituting leaves me with an extra root n in the denominator. nth term test shows that the nth term is zero, which doesn't really help. Any other ideas?
 May 14th, 2009, 07:52 AM #2 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Alternating Series $\sum \frac{ln(n)}{n\sqrt{n}}$ converges iff $\sum \frac{2^n*(ln(2^n)}{(2^n\sqrt{2^n})}$ converges which is $\sum \frac{nln2}{\sqrt{2^n}$ which converges by the root test. Therefore the series is absolutely convergent.
 May 14th, 2009, 08:22 AM #3 Newbie   Joined: Jun 2008 Posts: 20 Thanks: 0 Re: Alternating Series Hey, thanks for the reply. I must confess though, that I have no idea what you did. Could you break it down for me a little bit? Sorry
 May 14th, 2009, 02:35 PM #4 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Alternating Series if $a_n$ is a deacreasing positive sequence, at least eventually, then $\sum_{n}a_n$ converges and diverges with $\sum_{n}2^na_{2^n}$. This is sometimes called cauchy's condensation test.
 May 14th, 2009, 03:29 PM #5 Newbie   Joined: Jun 2008 Posts: 20 Thanks: 0 Re: Alternating Series I see. I've never seen that before... seems useful Thanks again, dman
 May 14th, 2009, 03:31 PM #6 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: Alternating Series most useful with logs and fractions.

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