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- Virginia
- George Mason University
- Mathematics
- Mathematics 213
- Kim
- tcu11_11_05.pdf

Sree Ram M.

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11.5 The Ratio and Root Tests 781 The Ratio and Root Tests The Ratio Test measures the rate of growth (or decline) of a series by examining the ratio For a geometric series this rate is a constant and the series converges if and only if its ratio is less than 1 in absolute value. The Ratio Test is a powerful rule extending that result. We prove it on the next page using the Comparison Test. ssar n + 1 d>sar n d = rd,gar n ,a n + 1 >a n . 11.5 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 781 Proof (a) Let r be a number between and 1. Then the number is positive. Since must lie within of when n is large enough, say for all In particular That is, These inequalities show that the terms of our series, after the Nth term, approach zero more rapidly than the terms in a geometric series with ratio More precisely, consider the series where for and Now for all n, and The geometric series converges because so con- verges. Since also converges. (b) From some index M on, The terms of the series do not approach zero as n becomes infinite, and the series diverges by the nth-Term Test. a n + 1 a n 7 1 and a M 6 a M + 1 6 a M + 2 6 Á . 1a n a n + 1 a n : r, P=r - rrR<1. 782 Chapter 11: Infinite Sequences and Series THEOREM 12 The Ratio Test Let be a series with positive terms and suppose that Then (a) the series converges if , (b) the series diverges if or is infinite, (c) the test is inconclusive if r = 1. rr 7 1 r 6 1 lim n: q a n + 1 a n = r. ga n 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 782 11.5 The Ratio and Root Tests 783 (c) The two series show that some other test for convergence must be used when In both cases, yet the first series diverges, whereas the second converges. The Ratio Test is often effective when the terms of a series contain factorials of ex- pressions involving n or expressions raised to a power involving n. EXAMPLE 1 Applying the Ratio Test Investigate the convergence of the following series. (a) (b) (c) Solution (a) For the series The series converges because is less than 1. This does not mean that 2 3 is the sum of the series. In fact, (b) If then and The series diverges because is greater than 1. (c) If then = 4sn + 1dsn + 1d s2n + 2ds2n + 1d = 2sn + 1d 2n + 1 : 1. a n + 1 a n = 4 n + 1 sn + 1d!sn + 1d! s2n + 2ds2n + 1ds2nd! # s2nd! 4 n n!n! a n = 4 n n!n!>s2nd!, r = 4 = s2n + 2ds2n + 1d sn + 1dsn + 1d = 4n + 2 n + 1 : 4. a n + 1 a n = n!n!s2n + 2ds2n + 1ds2nd! sn + 1d!sn + 1d!s2nd! a n + 1 = s2n + 2d! sn + 1d!sn + 1d! a n = s2nd! n!n! , a q n = 0 2 n + 5 3 n = a q n = 0 a 2 3 b n + a q n = 0 5 3 n = 1 1 - s2>3d + 5 1 - s1>3d = 21 2 . >r = 2>3 a n + 1 a n = s2 n + 1 + 5d>3 n + 1 s2 n + 5d>3 n = 1 3 # 2 n + 1 + 5 2 n + 5 = 1 3 # a 2 + 5 # 2 -n 1 + 5 # 2 -n b : 1 3 # 2 1 = 2 3 . g q n=0 s2 n + 5d>3 n , a q n = 1 4 n n!n! s2nd! a q n = 1 s2nd! n!n! a q n = 0 2 n + 5 3 n r = 1, For a q n = 1 1 n 2 : a n + 1 a n = 1>sn + 1d 2 1>n 2 = a n n + 1 b 2 : 1 2 = 1. For a q n = 1 1 n : a n + 1 a n = 1>sn + 1d 1>n = n n + 1 : 1. r = 1. a q n = 1 1 n and a q n = 1 1 n 2 R = 1. 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 783 Because the limit is we cannot decide from the Ratio Test whether the series converges. When we notice that we conclude that is always greater than because is always greater than 1. Therefore, all terms are greater than or equal to and the nth term does not ap- proach zero as The series diverges. The Root Test The convergence tests we have so far for work best when the formula for is rela- tively simple. But consider the following. EXAMPLE 2 Let Does converge? Solution We write out several terms of the series: Clearly, this is not a geometric series. The nth term approaches zero as so we do not know if the series diverges. The Integral Test does not look promising. The Ratio Test produces As the ratio is alternately small and large and has no limit. A test that will answer the question (the series converges) is the Root Test. n : q , a n + 1 a n = d 1 2n , n odd n + 1 2 , n even. n : q , = 1 2 + 1 4 + 3 8 + 1 16 + 5 32 + 1 64 + 7 128 + Á . a q n = 1 a n = 1 2 1 + 1 2 2 + 3 2 3 + 1 2 4 + 5 2 5 + 1 2 6 + 7 2 7 + Á ga n a n = e n>2 n , n odd 1>2 n , n even. a n ga n n : q . a 1 = 2, s2n + 2d>s2n + 1da n a n + 1 a n + 1 >a n = s2n + 2d>s2n + 1d, r = 1, 784 Chapter 11: Infinite Sequences and Series THEOREM 13 The Root Test Let be a series with for and suppose that Then (a) the series converges if (b) the series diverges if or is infinite, (c) the test is inconclusive if r = 1. rr 7 1 r 6 1, lim n: q 2 n a n = r. n Ú N,a n Ú 0ga n Proof (a) Choose an so small that Since the terms eventually get closer than to In other words, there exists an index such that 2 n a n 6 r +P when n Ú M. M Ú Nr.P 2 n a n 2 n a n : r,r +P61.P70R<1. 4100 AWL/Thomas_ch11p746-847 8/25/04 2:41 PM Page 784 11.5 The Ratio and Root Tests 785 Then it is also true that Now, a geometric series with ratio converges. By comparison, converges, from which it follows that converges. (b) For all indices beyond some integer M, we have so that for The terms of the series do not converge to zero. The series di- verges by the nth-Term Test. (c) The series and show that the test is not conclusive when The first series diverges and the second converges, but in both cases EXAMPLE 3 Applying the Root Test Which of the following series converges, and which diverges? (a) (b) (c) Solution (a) converges because (b) diverges because (c) converges because EXAMPLE 2 Revisited Let Does converge? Solution We apply the Root Test, finding that Therefore, Since (Section 11.1, Theorem 5), we have by the Sandwich Theorem. The limit is less than 1, so the series converges by the Root Test. lim n: q2 n a n = 1>22 n n : 1 1 2 ? 2 n a n ? 2 n n 2 . 2 n a n = e 2 n n>2, n odd 1>2, n even. ga n a n = e n>2 n , n odd 1>2 n , n even. B n a 1 1 + n b n = 1 1 + n : 0 6 1. a q n = 1 a 1 1 + n b n A n 2 n n 2 = 2 A2 n nB 2 : 2 1 7 1. a q n = 1 2 n n 2 B n n 2 2 n = 2 n n 2 2 n 2 n = A2 n nB 2 2 : 1 2 6 1. a q n = 1 n 2 2 n a q n = 1 a 1 1 + n b n a q n = 1 2 n n 2a q n = 1 n 2 2 n 2 n a n : 1. r = 1. g q n=1 s1>n 2 dg q n=1 s1>ndR = 1. n 7 M.a n 7 1 2 n a n 7 1,1

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