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 May 6th, 2013, 11:55 PM #1 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Convergence and divergence Hallo all! I am struggling with these questions, so a solution would be very helpful for my revision! (E is the sigma symbol) A) Infinity E = [(-6)^(m+1)]/[4^(5m)] n =0 B) Infinity E = [cos^2 (pi*n/]/[n sqrt(n)] n=0 C) infinity E = [4+cos(n)]/4^n n=0 please and thank you!
 May 7th, 2013, 12:58 AM #2 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Convergence and divergence Is the question whether the series diverge or converge? If so the first one can be shown to converge using the ratio test: $\sum_{n=0}^{\infty} \frac{(-6)^{n+1}}{4^{5n}}$ We need to find the limit: $\lim_{n\to\infty} \left| \frac{(-6)^{n+2}}{4^{5(n+1)}} \cdot \frac{(-6)^{n+1}}{4^{5n}}\right|= \lim_{n\to\infty} \left| \frac{-6}{4^5} \right| = \frac{3}{512}$ As this is less than 1 we can say the series converges.
 May 7th, 2013, 01:38 AM #3 Member   Joined: Jan 2012 Posts: 80 Thanks: 0 Re: Convergence and divergence For the second one the term is undefined at n=0. Did you mean: $\sum_{n=1}^{\infty} \frac{\cos^{2}(\frac{\pi n}{8})}{n\sqrt{n}}$
 May 7th, 2013, 02:03 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Convergence and divergence [color=#000000]1. $\overline{\lim_{n\to\infty}}\sqrt[n]{\left|\frac{(-6)^{n+1}}{4^{5n}}\right|}=\overline{\lim_{n\to\inf ty}}\sqrt[n]{\left|\frac{(-6)(-6)^{n}}{\left(4^{5}\right)^{n}}\right|}=\frac{|-6|}{4^{5}}\overline{\lim_{n\to\infty}}\sqrt[n]{|-6|}=\frac{6}{4^5}\cdot 1 =\frac{6}{4^5}<1=$ so the series converges. 2. $\sum_{n=1}^{\infty}\frac{\cos^{2}\left(\frac{\pi n}{8}\right)}{n\sqrt{n}}\leq \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}=\sum_{n=1}^ {\infty}\frac{1}{n^{\frac{3}{2}}}<\infty=$ so the series converges. 3. $\left|\sum_{n=0}^{\infty}\frac{4+\cos(n)}{4^n}\rig ht|\leq\sum_{n=0}^{\infty}\left|\frac{4+\cos(n)}{4 ^{n}}\right|\leq \sum_{n=0}^{\infty}\frac{4+1}{4^{n}}=\frac{20}{3}$ so the series converges.[/color]
 May 7th, 2013, 03:24 PM #5 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Re: Convergence and divergence thank you zardoz!
 May 9th, 2013, 09:24 PM #6 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Convergence and divergence Zardoz is a beast. It is worth pointing out however, that $\lim_{n\to+\infty}\sqrt[n]{x}=1$ is a nontrivial fact. If I were your grader, I would not give full credit. lol.
May 10th, 2013, 12:37 AM   #7
Math Team

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Re: Convergence and divergence

Quote:
 Originally Posted by guynamedluis Zardoz is a beast. It is worth pointing out however, that $\lim_{n\to+\infty}\sqrt[n]{x}=1$ is a nontrivial fact. If I were your grader, I would not give full credit. lol.
$\lim_{n\to\infty}\sqrt[n]{|x|}=\lim_{n\to\infty}\mathbb{e}^{\ln\left(\sqrt[n]{|x|}\right)}=\lim_{n\to\infty}\mathbb{e}^{\frac{1 }{n}\ln|x|}=\mathbb{e}^{0}=1$

May 10th, 2013, 12:59 AM   #8
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Re: Convergence and divergence

Quote:
 Originally Posted by guynamedluis It is worth pointing out however, that $\lim_{n\to+\infty}\sqrt[n]{x}=1$ is a nontrivial fact.
I feel the contrary : this is a trivial fact. As n increases, 1/n shrinks to 0 and the exponent too becomes 0. Hence, the expression tends 1.

 May 10th, 2013, 11:30 PM #9 Member   Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0 Re: Convergence and divergence Ooooh. It is indeed. I thought it was the sequence $\left\{\sqrt[n]{n}\right\}$. Of course, everything is trivial if you use logarithms and L'Hopital's rule.

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