My Math Forum Series Divergence Theorem
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 December 17th, 2011, 05:51 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Series Divergence Theorem I have a question about the divergence theorem that has confused me for some time. The divergence theorem says This seems to be contradicted in the example for a telescoping series below. Here they say that $s_n$ goes to 1, and the series sum is one.
 December 17th, 2011, 05:55 PM #2 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Series Divergence Theorem $\lim_{n \to \infty}\frac{1}{n(n+1)}=0$ I don't see the contradiction.
 December 17th, 2011, 05:59 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Series Divergence Theorem here
 December 17th, 2011, 06:02 PM #4 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: Series Divergence Theorem But $a_n=\frac{1}{n(n+1)}$ and as wnvl pointed out, $\lim_{n\to\infty}a_n=0$. However, $s_n=\sum_{i=1}^{n}a_i=$$1-\frac{1}{n+1}$$$ and so $\lim_{n\to\infty}s_n=1$. You are confusing $a_n$ with $s_n$.
 December 17th, 2011, 06:15 PM #5 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Series Divergence Theorem ok, I think I see now. $a_n$ is the sequence of terms. The Nth them must be zero otherwise you will never stop adding to the series, and it will have to diverge. $s_n$ is the partial sum of the first n-terms. It is confusing because all we are doing is putting $a_n$ in a another form and saying now it's now zero it's one.
 December 17th, 2011, 06:27 PM #6 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: Series Divergence Theorem Yes: $s_n=a_1+a_2+a_3+\cdots+a_{n-1}+a_n=\sum_{k=1}^na_k$ So, by the theorem cited above, we know the infinite series $s_{\infty}$ is divergent if $\lim_{n\to\infty}a_n\ne0$. This does not mean necessarily however, that the infinite series converges if $\lim_{n\to\infty}a_n=0$. Once we've established it diverges then we are done, but if the test for divergence fails, then we must prove whether it converges or not through the various tests for convergence.

 Tags divergence, series, theorem

Search tags for this page
,
,

,

,

,

,

,

,

,

,

,

,

,

,

# divergent sum and theorem

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post alan0354 Applied Math 0 August 10th, 2013 09:14 AM walter r Real Analysis 4 May 4th, 2013 07:34 PM patient0 Real Analysis 5 December 11th, 2010 06:17 AM everettjsj2 Calculus 3 February 28th, 2010 04:36 PM HammerTime Real Analysis 1 November 20th, 2008 04:18 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.