AP Calculus BC Review: Absolute and Conditional Convergence<\/a>.)<\/p>\nRatio Test.<\/strong> Let \nIf L<\/em> < 1, then the series converges absolutely.<\/li>\nIf L<\/em> > 1, then the series diverges.<\/li>\nHowever, if L<\/em> = 1, then the ratio test is inconclusive<\/strong>. (And then other test(s) would be needed to determine whether this series converges or diverges.)<\/li>\n<\/ul>\nThe Root Test<\/h3>\n The Root Test also serves to isolate a potential common ratio, but does so in a different way. This time, we take the n<\/em>th root of the general term and then allow n<\/em> to go to infinity.<\/p>\nThe inequalities are still the same, though. We still have to test whether the limit is less, greater, or equal to 1.<\/p>\n
Root Test.<\/strong> Let \nIf L<\/em> < 1, then the series converges absolutely.<\/li>\nIf L<\/em> > 1, then the series diverges.<\/li>\nHowever, if L<\/em> = 1, then the ratio test is inconclusive<\/strong>. (And then other test(s) would be needed to determine whether this series converges or diverges.)<\/li>\n<\/ul>\nSo how is the n<\/em>th root able to extract information about a potential common ratio? Let’s take a look at how the limit behaves when you actually have a geometric series, Σarn<\/sup><\/em>.<\/p>\n
The limit is L<\/em> = |r<\/em>|, which is the common ratio! (Well, technically, it’s the absolute value of the common ratio, but close enough.)<\/p>\nThree Useful Limits for Root Test<\/h4>\n It can be tricky to deal with a limit of the n<\/em>th root of some quantity. The following three properties will make your life a lot easier:<\/p>\n
Radius of Convergence<\/h3>\n The Ratio and Root Tests can help to find the radius of convergence for a series of functions. Here’s how it works:<\/p>\n
\nSet up the limit L<\/em> involving the general terms, as outlined above.<\/li>\nIf L<\/em> = 0, then the radius of convergence is infinite (R<\/em> = ∞).<\/li>\nIf L<\/em> = ∞, then the radius is zero (R<\/em> = 0).<\/li>\nOtherwise, if L<\/em> is nonzero and finite, then set up the inequality L<\/em> < 1, and solve for the expression |x<\/em> – c<\/em>|. The numerical value that shows up on the right side of the inequality is equal to R<\/em>.<\/li>\n<\/ol>\nExamples<\/h2>\nA Numerical Series<\/h3>\n Determine whether the following series converges or diverges.<\/p>\n
Your Intuition Might Deceive You<\/h4>\n When it comes to sequences and series, often our intuition can be misleading. For example, if you write out the first three terms of this series, you get:<\/p>\n
For n<\/em> = 0, we have 3(0!)\/(02<\/sup> + 1) = 3(1)\/1 = 3.<\/p>\nThen for n<\/em> = 1, we have 3(1!)\/(12<\/sup> + 1) = 3(1)\/2 = 1.5.<\/p>\nAnd for n<\/em> = 2, the value is 3(2!)\/(22<\/sup> + 1) = 3(2)\/5 = 1.2.<\/p>\nIt seems that the terms are just deceasing as n<\/em> increases, right? But the terms actually turn around and start increasing after a while!<\/p>\nSurprised cat is surprised.<\/figcaption><\/figure>\nFor n<\/em> = 3, we have 3(3!)\/(32<\/sup> + 1) = 3(6)\/10 = 1.8.<\/p>\nThen for n<\/em> = 4, the term is 3(4!)\/(42<\/sup> + 1) = 3(24)\/17 = 4.235.<\/p>\nIn fact, the terms themselves become unbounded as n<\/em> → ∞. That alone indicates that the series diverges (if you can prove it). <\/p>\nBut let’s see how Ratio or Root Test works in this case.<\/p>\n
Solution<\/h4>\n Usually when there are factorials in the general term, the Ratio Test is easier to apply than the Root Test.<\/p>\n
Now, because the limit is greater than 1 — and really, how much greater than 1 could you get than infinity!? — we conclude that this series diverges.<\/p>\n
When the Tests Fail<\/h3>\n Determine whether the alternating harmonic series converges or diverges.<\/p>\n
Solution<\/h4>\n You can pick either the Ratio or Root Test. I’ll set up the Ratio Test for you.<\/p>\n
This time, the limit is exactly 1. That means that the Ratio Test is inconclusive. It won’t help to try the Root Test instead — you’ll get the same result.<\/p>\n
So what do we do now?<\/p>\n
You’ll have to pick a different convergence test. I’d recommend the Alternating Series Test (AST), because this series is alternating, after all.<\/p>\n
Examining the absolute value of the general term, 1\/n<\/em>, we see that it does decrease down to zero (as n<\/em> → ∞). Therefore, by the AST, the alternating harmonic series converges.<\/p>\nFinding the Radius of Convergence<\/h3>\n Find the radius of convergence for the following series.<\/p>\n
Solution<\/h4>\n This one is just begging for Root Test!<\/p>\n
Since the limit was neither 0 nor infinity, we have to set the expression less than 1 and solve.<\/p>\n
The last line gives two pieces of information. Both are derived from the general template, |x<\/em> – c<\/em>| < R<\/em>, where c is the center, and R is the radius of convergence.<\/p>\n\nThe radius of convergence is R<\/em> = 1\/3. <\/li>\nThe center of the power series is c<\/em> = -2\/3.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"The Ratio and Root Tests are indispensable tools for finding out whether a series converges or diverges. These tests also play a large role in determining the radius and interval of convergence for a series of functions. In this article, we will discuss how the Ratio and Root Tests work. In addition, you’ll see a […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-10717","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\n
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