My Math Forum Comparison, Limit Comparison, and Splitting Sums.

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 April 7th, 2013, 01:09 PM #1 Newbie   Joined: Jan 2013 Posts: 26 Thanks: 0 Comparison, Limit Comparison, and Splitting Sums. I'm not really sure when each of these should be done. In fact, I don't really understand the reason that we use the limit comparison test. ?1/(n^2+1) So here I can simply say that P=2>1, so the original converges. ?1/N^3+N^2 Here, I would say that P=3>1, implying the original converges. But my solutions tell me that here I should use a limit comparison with 1/N^3, where the limit ---->N is = 1. So why did I do this? What did I prove by finding the limit that I didn't prove by just comparing directly? What is different in this that I had to use a limit comparison instead of direct? Just the added N^2? I don't understand why. And I also see that when I have some questions that have similar sums to the one directly above me, they just split the sum and evaluate the p series of both sums to find the divergence; is this an alternative to a limit comparison test as well? Thank you, guys.
 April 18th, 2013, 07:05 AM #2 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: Comparison, Limit Comparison, and Splitting Sums. [color=#000000]For the first one since $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{2}$, $\sum_{n=1}^{\infty}\frac{1}{n^2+1}<\sum_{n=1}^{\in fty}\frac{1}{n^2}<+\infty=$, so the series converges. For the second one here is another way, you can compute the result by doing the following $\sum_{n=1}^{\infty}\frac{1}{n^3+n^2}=\sum_{n=1}^{\ infty}\frac{1}{n^2}+\frac{1}{n+1}-\frac{1}{n}=\sum_{n=1}^{\infty}\frac{1}{n^2}-\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+1}}_{\text{telescoping series}}=\frac{\pi^2}{6}-\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=\frac{\pi^2}{6}-1$ .[/color]

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