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],"acf":[],"yoast_head":"\n
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