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 April 28th, 2012, 07:52 PM #1 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Convergence and Divergence's 1. $\sum_{n=1}^{\infty}$ $\frac{n^2}{3}$ Using the nth term test Diverges right? because $n^2 -> \infty$ (How do I do the arrows in latex?)
 April 28th, 2012, 08:03 PM #2 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Convergence and Divergence's Yes of course it diverges, if the sequence does not go to zero, it diverges. I think you can do an arrow with \rightarrow
 April 28th, 2012, 08:08 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 Math Focus: Calculus/ODEs Re: Convergence and Divergence's Use n\to\infty to get $n\to\infty$ The limit comparison test also shows it diverges.
 April 28th, 2012, 08:08 PM #4 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's So the $\sum_{n=1}^{\infty} \frac{2^n}{3}$ the r would be 2/3? I dont think that its right...
 April 28th, 2012, 08:11 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 Math Focus: Calculus/ODEs Re: Convergence and Divergence's You would have $r=2$: $r=\frac{\frac{2^{n+1}}{3}}{\frac{2^{n}}{3}}=2$ edit: Is that the series you meant to give in your first post...now your use of the root test makes more sense.
 April 28th, 2012, 08:12 PM #6 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's and the a would be one.
 April 28th, 2012, 08:14 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 Math Focus: Calculus/ODEs Re: Convergence and Divergence's You would have: $a=\frac{2}{3}$ $a=\frac{2^1}{3}=\frac{2}{3}$ You could write the series as: $\frac{2}{3}\cdot\sum_{n=0}^{\infty}2^n$
 April 28th, 2012, 08:16 PM #8 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's This one also converges? because i have a/1-r is -1/3<1
 April 28th, 2012, 08:17 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 Math Focus: Calculus/ODEs Re: Convergence and Divergence's This diverges because $r\ge1$. That formula works only for $r<1$.
 April 28th, 2012, 08:19 PM #10 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Convergence and Divergence's Then whats the deal with this formula... a/1-r? I thought i had to use that for convergence and divergence, but i can just go off of the r?

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