sign in sign up
<<
Home , Root test

Lecture Notes Ratio/Root Tests

Ratio Test & Root Test P (for an arbitrary-termed an)

an+1 For the , set ρ := lim . n→∞ an 1 pn note For the Root Test, set ρ := lim |an| = lim |an| n . n→∞ n→∞ Then X 0 ≤ ρ < 1 =⇒ an converges absolutely ρ = 1 =⇒ test is inconclusive (the test doesn’t tell us anything)

X th 1 < ρ ≤ ∞ =⇒ an diverges (by the n -term test for ) . 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 = |r| (1) an r r and so the ρ for the ratio test is

def an+1 by (1) ρ = lim = lim |r| = |r| n→∞ an n→∞ while for the root test we compute

1 n 1 A (n)( 1 ) A |an| n = |r | n = |r| n = |r| (2) and so the ρ for the root test is def 1 by (2) ρ = lim |an| n = lim |r| = |r| . n→∞ n→∞ P n So for a geometric::::::::::::::::series r , the ρ = |r| of both the ratio test and root test ::::big :::::time. We know  absolutely convergent when |r| < 1 X  rn is divergent when |r| = 1 ; ←− think of as the Borderline Case divergent when |r| > 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 ρ . Note that ρ ≥ 0.  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 from the ratio/root tests? Conditional convergence means P a converges while P |a | diverges. n n P Our ρ is computed using the |an| so the ratio/root tests can only tell us something about |an|. 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 (eg., n! or (2n)!) ...

because of :::::::::::::cancellation :::::::heaven (see next example). (2n)! Example 3. Simplify . Your final answer should not contain (!) 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

∞ 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→∞ an n→∞ [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) 3 (2 (n + 1) )! n+1 n+1 A (−1) 3 (2n)! = lim n n n→∞ (−1) 3 (2 (n + 1) )! n 1 A 3 3 (2n)! = lim (... now really in cancellation heaven - Example 3 will help) n→∞ 3n (2 (n + 1) )! A 1 1 = lim (3) = 3 lim = 3 · 0 = 0 < 1 . n→∞ (2n + 1) (2n + 2) n→∞ (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

∞ 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→∞ an n+1 2 A 3 n = lim n→∞ (n + 1)2 3n n+1 2 A 3 n = lim n→∞ 3n (n + 1)2 n 1 2 A 3 3   n  = lim n→∞ 3n (n + 1) 2 A  n  = lim (3) n→∞ (n + 1)  n 2  n 2 = 3 lim = 3 lim = 3 [1]2 = 3 > 1. n→∞ (n + 1) n→∞ (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/n 0 This follows from L’Hopital’s rule. Note that limn→∞ n is of the ∞ 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/n A 1 lim ln n1/n = lim =@ @∞ lim = lim = 0 n→∞ n→∞ n L’H n→∞ 1 n→∞ n   lim n1/n = lim exp ln n1/n = exp lim ln n1/n = exp (0) = 1 n→∞ n→∞ n→∞ or, if you like better,

ln n1/n −−−→n→∞ 0

1/n eln(n ) −−−→n→∞ e0 n1/n −−−→n→∞ 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

∞ 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 |an| = lim = lim = lim n→∞ n→∞ n2 n→∞ n2 n→∞ (n2)1/n A 3 1 A 1 1 = lim = 3 lim = 3 lim = 3 = 3 > 1. n→∞ n2/n n→∞ n2/n n→∞ (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

∞ X 23n+1 conditionally convergent nn n=1 divergent

23n+1 P Soln. Here an = nn ≥ 0. So |an| = an. Can an be conditionally convergent, which would be P P that an converges but |an| 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 |an| = lim = lim n→∞ n→∞ nn n→∞ (nn)1/n 3+1/n A 2 = lim n→∞ n 3 1/n A 2 2 = lim n→∞ n 1/n 3 2 = 2 lim (... lim 21/n = 20 = 1) n→∞ n n→∞ 3 1¡A 3 = 2 A¡ = = 2 · 0 = 0 < 1. ¡∞A 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