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 November 4th, 2011, 01:26 PM #1 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Comparison Test/Limit Comparison Test Questions Hi! I just had two questions on using the two Comparison tests. The first is the infinite sum starting with N=1 of.. 1 --- N! The second question is the infinite sum starting with N = 1 of... Sin(1/N) Thank you!
 November 4th, 2011, 01:35 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Comparison Test/Limit Comparison Test Questions Well, what exactly are the questions? To determine whether the sums converge or not? Since you title this "Comparison Test/Limit Comparison Test", what are some series you know that you can compare them with?
 November 4th, 2011, 01:41 PM #3 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Re: Comparison Test/Limit Comparison Test Questions Well, the book asks to determine whether the series converges or diverges. You are supposed to pick your own functions to compare it to. My problem is that for those two problems I can't think of how I would compare them with something that will tell me if they converge/diverge using those comparison tests.
 November 5th, 2011, 02:34 PM #4 Member   Joined: Jul 2009 Posts: 57 Thanks: 0 Re: Comparison Test/Limit Comparison Test Questions First off, they are decreasing and both bounded below by zero (lim n -> infinity nth term = 0.) For the first, it might be helpful to expand a little: $\sum_{n=1}^\infty \frac{1}{n!} = \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + ... + \frac{1}{1*2*3*4*5...n}$ Looking at the nth term you can see that just after 1*2 that each additional term is greater than 2. This means that: $\frac{1}{n!}$ Is eventually less than: $(\frac{1}{2})^n= \frac{1}{2^n}$ In fact, it is less when n > 3. Remember that it does not matter that 1/2^n is less than 1/n! for n <= 3 as this is only a finite amount. So by the comparison test: $\frac{1}{n!} < (\frac{1}{2})^n for n > 3 \sum_{n=3}^\infty \frac{1}{n!} < \sum_{n=3}^\infty (\frac{1}{2})^n$ Where $\sum_{n=3}^\infty (\frac{1}{2})^n$ Is a convergent geometric sequence So the fist series must be convergent. Use the same reason with the other one.

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