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 converges absolutely. 2
! 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 converges absolutely. (ln )
2 2 lim 2 2 = lim 2 2 2/ = lim 2 = 1 2
2 2 lim 2 2 = lim 2 2 2/ = lim 2 = 1 2 2 converges absolutely, 2
Harmonic and Alternating Harmonic Series Consider the harmonic series harmonic series ( 1) +1. 1 and the alternating Define the nth partial sum of the harmonic series as n 1 H n = and the n partial sum of the alternating n ( 1) +1 harmonic series as S n =.
Euler s Constant (1 of 2) 1.2 y 1.0 0.8 0.6 y=1/x a n =1/n 0.4 0.2 0.0 0 2 4 6 8 10 x Define the sequence {t n } n=1 as n 1 n t n = 1 x dx + 1 n = ln n + 1 > 0.
Euler s Constant (2 of 2) {t n } n=1 is a decreasing sequence since ( ) n 1 t n t n+1 = ln n + n+1 1 ln(n + 1) + = ln(n + 1) ln n 1 ( n + 1 = ln 1 + 1 ) 1 n n + 1 > 0. Since t n > 0 (bounded below) then lim n t n = γ (called Euler s constant). H n ln n γ as n.
Alternating Harmonic Series Sum (1 of 2) Claim: S 2n = H 2n H n. Proof. H 2n H n = H 2n 1 2 H n 1 2 H n = 1 + 1 2 + 1 3 + + 1 2n 1 + 1 = 1 + 1 3 + + 1 2n 1 ( 1 2n ( 1 2 + 1 4 + + 1 2n = 1 1 2 + 1 3 1 4 + + 1 2n 1 1 2n = S 2n 2 + 1 4 + + 1 2n ) ) 1 2 H n
Alternating Harmonic Series Sum (2 of 2) lim S 2n = lim (H 2n H n ) n n = γ + ln(2n) (γ + ln n) = ln(2n) ln n = ln 2n n = ln 2 Thus we have proved the Alternating Harmonic Series converges to ln 2. ( 1) +1 = ln 2
Rearrangements of Series Rearranging the order of terms in a finite sum leaves the sum unchanged. Rearranging the order of terms in an absolutely convergent infinite sum leaves the sum unchanged. Any conditionally convergent infinite sum can be rearranged to give a different sum.
Example: Alternating Harmonic Series ln 2 = 1 1 2 + 1 3 1 4 + 1 5 1 6 + 1 7 1 8 + 1 2 ln 2 = 1 2 1 4 + 1 6 1 8 + 1 10 1 12 + 1 14 1 16 + = 0 + 1 2 + 0 1 4 + 0 + 1 6 + 0 1 8 + 0 + 1 10 + 0 1 12 + 0 + 1 14 + 0 1 16 + Add the first and third equations. 3 2 ln 2 = 1 + 1 3 1 2 + 1 5 + 1 7 1 4 +
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