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Lecture Notes Ratio/Root Tests Ratio Test & Root Test (For an Arbitrary Lecture Notes Ratio/Root Tests Ratio Test & Root Test P (for an arbitrary-termed series an) an+1 For the Ratio Test, set ρ := lim : n!1 an 1 pn note For the Root Test, set ρ := lim janj = lim janj n : n!1 n!1 Then X 0 ≤ ρ < 1 =) an converges absolutely ρ = 1 =) test is inconclusive (the test doesn't tell us anything) X th 1 < ρ ≤ 1 =) an diverges (by the n -term test for divergence) . The symbol ρ, pronounced rho, is the Greek small cases letter r and is used for a reason ::: Idea behind (ie. how to remember) the Ratio Test & Root Test For a ::::::::::geometric::::::series X n n r (so an = r ) for the ratio test we compute n+1 n 1 an+1 r A r r A = n = n = jrj (1) an r r and so the ρ for the ratio test is def an+1 by (1) ρ = lim = lim jrj = jrj n!1 an n!1 while for the root test we compute 1 n 1 A (n)( 1 ) A janj n = jr j n = jrj n = jrj (2) and so the ρ for the root test is def 1 by (2) ρ = lim janj n = lim jrj = jrj : n!1 n!1 P n So for a geometric::::::::::::::::series r , the ρ = jrj of both the ratio test and root test ::::big :::::time. We know 8 absolutely convergent when jrj < 1 X <> rn is divergent when jrj = 1 ; − think of as the Borderline Case :>divergent when jrj > 1 : Basically, if your ρ of the ratio/root test stays away from the Borderline Case, P P n then a given series an behaves like the geometric series ρ . Note that ρ ≥ 0. 8 absolutely convergent when 0 ≤ ρ < 1 X <> an is test is inconclusive when ρ = 1 ; − the Borderline Case :>divergent when ρ > 1 : Question 1. Can you conclude conditional convergence from the ratio/root tests? Conditional convergence means P a converges while P ja j diverges. n n P Our ρ is computed using the janj so the ratio/root tests can only tell us something about janj. So the ratio/root tests cannot tell us that a series is conditionally convergent. Prof. Girardi/19.01.01 (yr.mn.dy) Page 1 of 4 Series Lecture Notes Ratio/Root Tests Remark 2. Often we use the ::::::Ratio:::::Test if have factorials (eg., n! or (2n)!) ::: because of :::::::::::::cancellation :::::::heaven (see next example). (2n)! Example 3. Simplify . Your final answer should not contain factorial (!) signs. (2 (n + 1) )! (2n)! A (2n)! A 1 · 2 ··· (2n) = = (2 (n + 1) )! (2n + 2)! 1 · 2 ··· (2n + 2) to better see cancellation, let's write out some more terms A 1 · 2 ··· (2n) = 1 · 2 ··· (2n) (2n + 1) (2n + 2) we can also just view this as A (2n)! A 1 = = : (2n)! (2n + 1) (2n + 2) (2n + 1) (2n + 2) A I. Key Idea 4. For easier cancellation, express (2n + 2)! = (2n)! (2n + 1) (2n + 2). Example 5. (AC/CC/Divg) Below the choice-boxes , :::::::::carefully :::::::justify the given series's behavior. Be sure to specify which test(s) you are using and clearly explain your logic. Then check the correct choice-box. absolutely convergent 1 n X (−1) 3n conditionally convergent (2n)! n=1 divergent (−1)n3n Soln: Here an = (2n)! . (Be sure to say what you are thinking.) Foreseeing cancellation in an+1 , let's try the ratio test. an ::::::::: n+1 n+1 a A (−1) 3 (2n)! ρ = lim n+1 = lim n n n!1 an n!1 [2 (n + 1)]! (−1) 3 group together like terms to get into cancellation heaven n+1 n+1 A (−1) 3 (2n)! = lim (::: next get rid of unneeded |·|) n n n!1 (−1) 3 (2 (n + 1) )! n+1 n+1 A (−1) 3 (2n)! = lim n n n!1 (−1) 3 (2 (n + 1) )! n 1 A 3 3 (2n)! = lim (::: now really in cancellation heaven - Example 3 will help) n!1 3n (2 (n + 1) )! A 1 1 = lim (3) = 3 lim = 3 · 0 = 0 < 1 : n!1 (2n + 1) (2n + 2) n!1 (2n + 1) (2n + 2) (−1)n3n P (go up and check the \abs. convergent" box) Since ρ < 1, by the :::::ratio::::test, (2n)! is absolutely convergent. a n I. Observation 6. Look at what taking the absolute value of n+1 did to the (−1) ::: basically an could ignore the (−1)n ::: when computing the an+1 . an Prof. Girardi/19.01.01 (yr.mn.dy) Page 2 of 4 Series Lecture Notes Ratio/Root Tests Example 7. (AC/CC/Divg) Below the choice-boxes , :::::::::carefully :::::::justify the given series's behavior. Be sure to specify which test(s) you are using and clearly explain your logic. Then check the correct choice-box. absolutely convergent 1 n X (−1) 3n conditionally convergent n2 n=1 divergent n n (−1) 3 an+1 Soln: Here, a = 2 . The ratio test will lead to cancellation heaven when we compute . n n an an+1 ρ = lim n!1 an n+1 2 A 3 n = lim n!1 (n + 1)2 3n n+1 2 A 3 n = lim n!1 3n (n + 1)2 n 1 2 A 3 3 n = lim n!1 3n (n + 1) 2 A n = lim (3) n!1 (n + 1) n 2 n 2 = 3 lim = 3 lim = 3 [1]2 = 3 > 1: n!1 (n + 1) n!1 (n + 1) (−1)n3n P (go up and check the \divergent" box) Since ρ > 1, by the :::::ratio:::::test, n2 is divergent. Remark 8. Useful for the Root Test is one of our Commonly Occurring Limits of Sequences from the section on sequence: lim n1=n = 1: (8.1) n!1 1=n 0 This limit follows from L'Hopital's rule. Note that limn!1 n is of the indeterminate form 1 so we take the natural log of n1=n, then use L'Hopital's rule, then exponentiate back. 1 ln n ln n1=n = ln n = n n ln n 1 1=n A 1 lim ln n1=n = lim =@ @1 lim = lim = 0 n!1 n!1 n L'H n!1 1 n!1 n lim n1=n = lim exp ln n1=n = exp lim ln n1=n = exp (0) = 1 n!1 n!1 n!1 or, if you like better, ln n1=n −−−!n!1 0 1=n eln(n ) −−−!n!1 e0 n1=n −−−!n!1 1 : Prof. Girardi/19.01.01 (yr.mn.dy) Page 3 of 4 Series Lecture Notes Ratio/Root Tests Example 9. (AC/CC/Divg) Below the choice-boxes , :::::::::carefully :::::::justify the given series's behavior. Be sure to specify which test(s) you are using and clearly explain your logic. Then check the correct choice-box. absolutely convergent 1 n X (−1) 3n conditionally convergent n2 n=1 divergent Soln. Hey, this is Example 7, which we did using the ratio test. (−1)n3n Well, let's see how it goes using the :::::root ::::test. Here, an = n2 . n n 1=n n 1=n n 1=n 1=n (−1) 3 A 3 A (3 ) ρ = lim janj = lim = lim = lim n!1 n!1 n2 n!1 n2 n!1 (n2)1=n A 3 1 A 1 1 = lim = 3 lim = 3 lim = 3 = 3 > 1: n!1 n2=n n!1 n2=n n!1 (n1=n)2 (1)2 (−1)n3n P (go up and check the \divergent" box) Since ρ > 1, by the :::::root ::::test, n2 is divergent. Example 10. (AC/CC/Divg) Below the choice-boxes , :::::::::carefully :::::::justify the given series's behavior. Be sure to specify which test(s) you are using and clearly explain your logic. Then check the correct choice-box. absolutely convergent 1 X 23n+1 conditionally convergent nn n=1 divergent 23n+1 P Soln. Here an = nn ≥ 0. So janj = an. Can an be conditionally convergent, which would be P P that an converges but janj diverges? No way since an ≥ 0. Let's go up and indicate that by crossing out conditionally convergent and writing \cannot be since it's a positive termed series". Since we have all those n's in the exponent, let's try the root::::::::test. 3n+1 1=n 3n+1 1=n 1=n 2 A (2 ) ρ = lim janj = lim = lim n!1 n!1 nn n!1 (nn)1=n 3+1=n A 2 = lim n!1 n 3 1=n A 2 2 = lim n!1 n 1=n 3 2 = 2 lim (::: lim 21=n = 20 = 1) n!1 n n!1 3 1¡A 3 = 2 A¡ = = 2 · 0 = 0 < 1: ¡1A 23n+1 P (go up and check the \abs. convergent" box) Since ρ < 1, by the :::::ratio:::::test, nn is absolutely convergent. Remark 11. If trying Ratio/Root test and get ρ = 1, then the test is inconclusive and we need to try a different test. Prof. Girardi/19.01.01 (yr.mn.dy) Page 4 of 4 Series.
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