Absolute Convergence and the Ratio Test MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018
Bacground Remar: All previously covered tests for convergence/divergence apply only to positive term series (except for the Alternating Series Test).
Bacground Remar: All previously covered tests for convergence/divergence apply only to positive term series (except for the Alternating Series Test). Question: what about series which do not consist exclusively of positive terms, but which are not alternating series?
Bacground Remar: All previously covered tests for convergence/divergence apply only to positive term series (except for the Alternating Series Test). Question: what about series which do not consist exclusively of positive terms, but which are not alternating series? Example cos 2 = cos 1 + cos 2 4 + cos 3 9 + 0.540302 0.104037 0.109999 +
Absolute Convergence Definition An infinite series a is absolutely convergent if the series converges. a = a 1 + a 2 + a 3 +
Absolute Convergence Definition An infinite series a is absolutely convergent if the series converges. a = a 1 + a 2 + a 3 + Remar: The series a is a positive term series.
Conditional Convergence Definition An infinite series a is conditionally convergent if the series converges but the series a diverges.
Conditional Convergence Definition An infinite series a is conditionally convergent if the series converges but the series For an arbitrary series a diverges. a, the series may be classified in only one of the following ways: absolutely convergent conditionally convergent divergent
Examples Determine which of the following infinite series are absolutely convergent, conditionally convergent, or divergent. cos 1. 2 2. 3. 4. ( 1) +1 1 ( 1) tan 1 3 sin( π/6)
cos 2 cos 2 1 for all = 1, 2,.... 2
cos 2 cos 2 1 for all = 1, 2,.... 2 1 The series converges (p-series Test) which implies 2 cos 2 converges.
cos 2 cos 2 1 for all = 1, 2,.... 2 1 The series converges (p-series Test) which implies 2 Therefore, cos 2 cos 2 converges. converges absolutely.
( 1) +1 1 1 ( 1)+1 1 for all = 1, 2,....
( 1) +1 1 1 ( 1)+1 1 for all = 1, 2,.... 1 The series diverges (harmonic series).
( 1) +1 1 1 ( 1)+1 1 for all = 1, 2,.... 1 The series diverges (harmonic series). The series ( 1) +1 1 converges by the Alternating Series Test.
( 1) +1 1 1 ( 1)+1 1 for all = 1, 2,.... 1 The series diverges (harmonic series). The series ( 1) +1 1 converges by the Alternating Series Test. Therefore, ( 1) +1 1 converges conditionally.
( 1) tan 1 3 ( 1) tan 1 3 π/2 for all = 1, 2,.... 3
( 1) tan 1 3 ( 1) tan 1 3 π/2 for all = 1, 2,.... 3 1 The series converges (p-series Test) which implies 3 π/2 3 converges.
( 1) tan 1 3 ( 1) tan 1 3 π/2 for all = 1, 2,.... 3 1 The series converges (p-series Test) which implies 3 π/2 3 converges. By the Comparison Test ( 1) tan 1 3 converges.
( 1) tan 1 3 ( 1) tan 1 3 π/2 for all = 1, 2,.... 3 1 The series converges (p-series Test) which implies 3 By the Comparison Test Therefore, π/2 3 ( 1) tan 1 3 converges. ( 1) tan 1 3 converges. converges absolutely.
sin( π/6) sin( π/6) 1 for all = 1, 2,.... 3/2
sin( π/6) sin( π/6) 1 for all = 1, 2,.... 3/2 1 The series converges (p-series Test) which 3/2 implies sin( π/6) converges.
sin( π/6) sin( π/6) 1 for all = 1, 2,.... 3/2 1 The series converges (p-series Test) which 3/2 implies sin( π/6) converges. Therefore, sin( π/6) converges absolutely.
Absolute Convergence Implies Convergence Theorem If a converges then a converges.
Absolute Convergence Implies Convergence Theorem If a converges then a converges. Proof. a a a 0 a + a 2 a Therefore (a + a ) converges by the Comparison Test. (a + a a ) = a converges.
Ratio Test Theorem (Ratio Test) Given a, with a 0 for all, suppose that Then lim a +1 a = L. 1. if L < 1, the series converges absolutely, 2. if L > 1, the series diverges, 3. if L = 1, there is no conclusion.
Examples Use the Ratio Test to determine the convergence or divergence of the following series. 20 1. 2 2. 3. 4.! 1 1 2
20 2 lim (+1) 20 2 +1 20 2 2 = lim 2 +1 1 = lim 2 ( + 1 ( 1 + 1 ) 20 ) 20 = 1 2 < 1
20 2 lim (+1) 20 2 +1 20 2 2 = lim 2 +1 1 = lim 2 ( + 1 ( 1 + 1 ) 20 ) 20 = 1 2 < 1 20 2 converges absolutely.
! lim (+1) +1 (+1)!!! ( + 1) +1 = lim ( + 1)! = lim = lim 1 ( + 1) +1 + 1 ( + 1) ( = lim 1 + 1 ) = e > 1
! lim (+1) +1 (+1)!!! ( + 1) +1 = lim ( + 1)! = lim = lim 1 ( + 1) +1 + 1 ( + 1) ( = lim 1 + 1 ) = e > 1! diverges.
1 lim 1 +1 1 = lim + 1 = 1 The Ratio Test reaches no conclusion. The series is recognized as the harmonic series and therefore diverges.
1 2 lim 1 (+1) 2 1 2 = lim 2 ( + 1) 2 = 1 The Ratio Test reaches no conclusion. The series is recognized as a convergent p-series and therefore converges absolutely.
Root Test Theorem (Root Test) Given a, suppose that Then lim a = L. 1. if L < 1, the series converges absolutely, 2. if L > 1, the series diverges, 3. if L = 1, there is no conclusion.
Examples Use the Root Test to determine the convergence or divergence of the following series. (ln ) /2 1. 2. 3. =2 2 +1 (ln ) 2 2
(ln ) /2 lim (ln ) /2 (ln ) = lim /2 = lim (ln ) 1/2 = 0 < 1
(ln ) /2 lim (ln ) /2 (ln ) = lim /2 (ln ) 1/2 = lim = 0 < 1 (ln ) /2 converges absolutely.
2 +1 (ln ) =2 lim 2 +1 (ln ) = lim 2 +1 (ln ) = lim 2 1+1/ ln = 0 < 1
2 +1 (ln ) =2 lim 2 +1 (ln ) = lim 2 +1 (ln ) 2 1+1/ = lim ln = 0 < 1 =2 2 +1 (ln ) converges absolutely.
2 2 lim 2 2 = lim 2 2 = lim 2/ 2 = 1
2 2 lim 2 2 The Root Test is inconclusive. according to the Ratio Test. = lim 2 2 2/ = lim 2 = 1 2 2 converges absolutely,
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