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 April 28th, 2012, 10:17 PM #21 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Convergence and Divergence's We just need for the magnitude of each successive term to decrease at a certain critical rate or beyond to get a convergent series, regardless of how large the first terms are.
April 29th, 2012, 05:46 PM   #22
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Re: Convergence and Divergence's

Quote:
 Originally Posted by MarkFL It in fact converges to $\frac{\pi^2}{2}\approx4.9348$ .

Is this the $a_n+1/a_n$ formula?

 April 29th, 2012, 07:31 PM #23 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Convergence and Divergence's No, that formula (I think you mean $\frac{a_{n+1}}{a_n}$) is used in the ratio test, which was inconclusive on the series. I found the sum using the computer...I bet our friend Zardoz could show how to manually find it.
 April 29th, 2012, 07:37 PM #24 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's You know about him?
 April 29th, 2012, 07:42 PM #25 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Convergence and Divergence's Yes, our resident analyst and I go waaaay back.
 April 29th, 2012, 07:43 PM #26 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's It really is a boys club, aye?
 April 29th, 2012, 07:50 PM #27 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Convergence and Divergence's Unfortunately...yes. :P
April 30th, 2012, 08:03 PM   #28
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Re: Convergence and Divergence's

Quote:
 Originally Posted by FreaKariDunk 1. $\sum_{n=1}^{\infty}$ $\frac{n^2}{3}$
Will you show me the limit comparison test?

I cannot seem to follow it in the book

 April 30th, 2012, 08:22 PM #29 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Convergence and Divergence's By simple inspection we can tell that diverges, and the use of the limit comparison test would really be redundant. $\sum_{n=1}^{\infty}\frac{n^2}{3}=\lim_{n\to\infty} $$\sum_{k=1}^{n}\frac{k^2}{3}$$=\lim_{n\to\infty}\ (\frac{n(n+1)(2n+1)}{18}\)=\infty$

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