Math 104 – Rimmer 12.6 Absolute Convergence and the Ratio and Root Tests An infinite series ∞ ∞ ∑an is called absolutely convergent if the positive series ∑ a n converges. n=1 n = 1 Absolute convergence implies converges. (If the series of absolute value converge s, then the original series also converg es ) If the series of absolute value diverges , it is still possible for the original series to converge. Use the Alternating Series Test on the o riginal series. If the Alternating Series Test gives con vergence, then this is a special type of convergence. An infinite series ∞ ∞ ∑an is called conditionally convergent if it converges but ∑ a n diverg es . n=1 n = 1 Math 104 – Rimmer 12.6 Absolute Convergence and the Ratio and Root Tests ∞ convergent ∑an is absolutely convergent n=1 ∞ ∞ ∑an is convergent n=1 ∑ an n=1 divergent ∞ On ∑ an try : n=1 a) the Alternating Series Test, or b) the Test for Divergence If the Alternating Series Test gives con vergence, If the Test for Divergence gives diverge nce, ∞ ∞ ∑an is conditionally convergent ∑an is divergent n=1 n=1 Math 104 – Rimmer 12.6 Absolute Convergence and the Ratio and Root Tests A major difference between absolutely co nvergent and conditionally convergent comes in the rearrangement of the terms. ∞ If a is absolutely convergent with sum s , ∑ n ∞ n=1 then any rearrangement of the sum ∑an will have the same sum s . n=1 ∞ If a is conditionally convergent and r is any real number, ∑ n ∞ n=1 then there is a rearrangement of the sum ∑an that has the sum r . n=1 n−1 ∞ ()−1 1 1 1 1 1 1 11111 1 ∑ =−+−+−+1 = ln 2 ()We will show this later 1−+−+−+ = ln 2 n=1 n 2 3 4 5 6 2 23456 2 111 1 1111111 − + + = 1−+−+−+−+ = ln 2 ln 2 2345678 246 2 1 1 1 1 111 1 1 0+ +−+++ 0 0 = ln 2 + 0 ++−++++−+ 0 0 0 = ln2 2 4 6 2 246 8 2 111 11 3 1+ −++ −+= ln 2 different sums 325 74 2 same terms Math 104 – Rimmer Determine whether the series is absolute ly convergent, 12.6 Absolute Convergence and the Ratio and Root Tests conditionally convergent, or divergent. n+1 n+1 ∞ cos (nπ ) ∞ ()−1 ∞ ()−4 i) ii iii ∑ 3 ) ∑ ) ∑ n n=1 n n=1 n n=1 3 ∞ ∞ ∞ 1 1 ∞ ∞ ∞ ∞ ∞ n+1 ∞ n = 1 1 4 4 ∑an ∑ = ∑ 3/ 2 3 ∑an = ∑ = ∑ 1/ 2 ∑an = ∑ = 4 n=1 n = 1 n n=1 n n ∑ n=1 n = 1 n n=1 n n=1 n = 1 3 n=1 3 convergent p − series divergent p − series divergent geom. series ∞ cos ()nπ n+1 n+1 ∞ ⇒ ∑ is ∞ −4 3 (−1) ( ) = use T. F . D .: n 1 n ∑ use A. S . T .: ∑ n n=1 3 absolutely convergent n=1 n 1 n+1 = ()−4 bn is decreasing, lim does not exist n →∞ n n n+1 3 (−4) limn = −∞ , for n even and limbn = 0 n→∞ 3 n→∞ n+1 ()−4 lim= ∞ , for n odd convergent by A. S . T . n→∞ 3n n+1 n+1 ∞ ()−1 ∞ ()−4 ⇒ ⇒ ∑ is ∑ n is n=1 n n=1 3 conditionally convergent divergent Math 104 – Rimmer 12.6 Absolute Convergence and the Ratio and Root Tests an+1 Let {}an be a sequence and assume that the following limit exists: lim = L n→∞ an ∞ i) If L<1, then the series ∑ a n is absolutely c onvergent. n=1 ∞ ii) If L >1 or if the limit is infinite, then the series ∑a n is divergent. n=1 iii) If L =1, the Ratio Test is inconclusive. (the series could be absolutely convergen t, conditionally convergent, or divergen t) Math 104 – Rimmer 12.6 Absolute Convergence and the Ratio and Root Tests n Let {an } be a sequence and assume that the following limit exists: lim an = L n→∞ ∞ i) If L<1, then the series ∑ a n is absolutely c onvergent. n=1 ∞ ii) If L >1 or if the limit is infinite, then the series ∑a n is divergent. n=1 iii) If L =1, the Root Test is inconclusive. (the series could be absolutely convergen t, conditionally convergent, or divergen t) Math 104 – Rimmer 12.6 Absolute Convergence Determine whether the series is converge nt or divergent. and the Ratio and Root Tests ∞ 3 n ∞ n! i) ∑ n ii ) 4 ∑ n n=1 n=1 4 3 ()n +1 3 1 n 3 n n+1 ()n +1 4 ()n +1 1 an+1 4 = ⋅ 4 i) lim= lim lim 3n+ 1 = lim ⋅ = n→∞ n →∞ 3 n→∞ n 4 n→∞ 3 n an n n 4⋅ 4 4 4n ∞ n3 ∑ n is convergent n=1 4 (n +1) ! a n+1 (n +1) ! 4n ()n+1 n ! 4n n +1 ii n+1 = 4 ) lim lim =lim ⋅ n+1 = lim ⋅ = lim = ∞ n→∞ n →∞ n! n→∞ n→∞ n n→∞ an n! 4 n! 4⋅ 4 4 4n ∞ n! ∑ n is divergent n=1 4 Math 104 – Rimmer 12.6 Absolute Convergence Determine whether the series is converge nt or divergent. and the Ratio and Root Tests n ∞ ()2n + 1 ∞ n2 iii ) ∑ 2n iv ) ∑ n=1 n n=1 ()2n + 1 ! n n 1/ n ()2n + 1 2n + 1 2n + 1 n = lim = lim = 0 iii) lim a n = lim n n 2 →∞ 2 n→∞ n →∞ 2 n→∞ n n n ()n deg(num .) < deg.( denom . ) n ∞ (2n + 1 ) ∑ 2n is convergent n=1 n 2 ()n +1 2n + 1 + 1! 2 an+1 ()[] n+1 2 n + 1! iv ) lim= lim () () n→∞a n →∞ n2 =lim ⋅ n n→∞ n2 ()2 n + 3 ! ()2n + 1 ! ∞ n2 ∑ is convergent n=1 ()2n + 1 ! 1 2 ()n +1 ()2n + 1 ! 1 = lim 2 ⋅ = lim n→∞ n ()()()2n+⋅ 3 2 n + 22 n + 1! n→∞ ()()2n+ 3 ⋅ 2 n + 2 = 0.