MAT 127: Calculus C, Fall 2010Notes on the Ratio Test: for Sequences and Series0 IntroductionIn the textbook, the Ratio Test is introduced in Section 8.4 as one of many convergence tests forseries∞Xn=1an= a1+ a2+ . . . .A convergence test for sequences a1, a2, . . . can be deduced from RT for Series using the DivergenceTest for Series (box 6 or 7 on p570). However, this convergence test can also be obtained moredirectly. Just like RT for Series, RT for Sequences works well for some sequences and is useless forothers. On the other hand, RT for Series is perhaps the most frequently used convergence test forseries, while RT for Sequences is perhaps the least frequently used convergence test for series (butit does work very nicely in some cases!).In order to use either of the ratio tests, look at the ratio of the absolute values of two consecutiveterms, |an+1|/|an|; note that the higher-numbered term goes to the top of this fraction and thelower-numbered term goes to the bottom. For example, for the sequence an= (−1)n7n/n!, thisratio is|an+1||an|=7n+1/(n+1)!7n/n!=7n+17n·n!(n+1)!=7n· 77n·n!n! · (n+1)=7n+1.This produces a new sequence a2/a1, a3/a2, a4/a3, . . ., which consists of non-negative terms.1 Ratio Test for SequencesFirst, suppose we want to determine whether the sequence a1, a2, . . . has a limit. We could insteadlook at the sequence of the absolute values of the ratios of consecutive terms |an+1|/|an|. If thelatter sequence has a limit L (which must necessarily be non-negative) andL ≡ limn−→∞|an+1||an|< 1, then limn−→∞an= 0. (1)On the other hand, ifL ≡ limn−→∞|an+1||an|> 1 or|an+1||an|−→ ∞, then |an| −→ ∞, (2)and so the sequence andiverges (or possibly “converges” to infinity). Finally, ifL ≡ limn−→∞|an+1||an|= 1, then this test says nothing. (3)In this last case, you’ll need to find some other way to determine if the sequence a1, a2, . . . converges.For example, for the sequence an= (−1)n7n/n!,limn−→∞|an+1||an|= limn−→∞7n + 1=7∞ + 1= 0.Since 0 < 1, this sequence converges to 0. On the other hand, for the sequence an= 2n/n,|an+1||an|=2n+1/(n+1)2n/n=2n+12n·nn+1=2n· 22n·1n/n + 1/n= 211 + 1/n−→ 211 + 1/∞= 211 + 0= 2.Since 2 > 1, this sequence diverges (actually “converges” to ∞).For the sequence 1, 1, 1, . . ., the limit of the ratios of the absolute values of consecutive terms is 1(all of these ratios are 1) and this sequence converges 1. For the sequence −1, 1, −1, 1 . . ., the limitof the ratios of the absolute values of consecutive terms is also 1 (all of these ratios are again 1), butthis sequence diverges (it keeps on jumping between 1 and -1). This shows that RT for Sequencesis useless in the caseL ≡ limn−→∞|an+1||an|= 1.Even a sequence a1, a2, a3, . . . with L=1 and an>0 need not converge. For example, leta1= 211, a2= 211+12, a3= 211+12+13, . . . . (4)Then,|an+1||an|=211+12+...+1n+1n+1211+12+...+1n=211+12+...+1n· 21n+1211+12+...+1n= 21n+1−→ 21∞+1= 20= 1.However,211+12+...+1n−→ 2∞= ∞,because the harmonic series∞Xn=11n=11+12+13+ ...diverges by the p-Series Test on p578 (this is p = 1; see also Example 7 on p569). Thus, thesequence of positive terms a1, a2, . . . described in (4) diverges (it actually “converges” to ∞), eventhough L=1 in this case.Why is (1) true? The assumption in (1) is that the ratios |an+1|/|an| get very close to L as nincreases. Since L<1 in this case, this means that |an+1|/|an| < (L+1)/2 if n is very large, so that|an+1| <L+12|an|for all n larger than some N. Thus,aN+n<L+12aN+n−1<L+12·L+12aN+n−2. . . <L+12· . . . ·L+12| {z }naN=L+12n|aN|.2Thus, the sequence |aN|, |aN+1|, . . . is squeezed between the sequence 0, 0, . . . and the geometricsequence|aN|,L+12|aN|,L+122|aN|, . . .This geometric sequence converges to 0 because |(L+1)/2|<1 in this case (see box 7 on p560). Thus,the sequence |aN|, |aN+1|, . . . also converges to 0 by the Squeeze Theorem for Sequences on p557and so does the sequence aN, aN+1, . . . by Theorem 4 on p557. Since the convergence of a sequencehas nothing to do with how it begins, it follows that the original sequence a1, a2, . . . also converges.Why is (2) true? The assumption in (2) is that the ratios |an+1|/|an| become larger than somenumber r > 1 as n increases (in the first case, we can take r = (L+1)/2; in the second case, r canbe taken to be any number larger than 1). Thus, |an+1|>r|an| for all n larger than some N and soaN+n> raN+n−1> r · raN+n−2. . . > r ·. . . ·r| {z }naN= rn|aN|.So, the terms in the sequence |aN|, |aN+1|, . . . are larger than the terms in the geometric sequence|aN|, r|aN|, r2|aN|, . . .This geometric sequence diverges (actually “converges” to ∞) because r > 1 (see box 7 on p560).Thus, the sequence |aN|, |aN+1|, . . . also diverges (also “converges” to ∞), since its terms are evenlarger. Since the convergence of a sequence has nothing to do with how it begins, it follows thatthe original sequence a1, a2, . . . also diverges.Because of (1), (2), and (3), RT for Sequences can detect convergence of sequences a1, a2, . . . withlimit 0 only (and even of only some of these). So it can rarely be used to detect convergentsequences, but whenever it is applicable, RT for Sequences determines the limit of convergentsequences immediately. RT for Sequences has a good chance of working for sequences that involvefactorials and powers n (e.g. n!, 3n, nn), but has no chance of working for sequences that involveonly powers of n (e.g. n3).2 Ratio Test for SeriesRT for Series works similarly to RT for Sequences to determine whether certain series∞Xn=1an= a1+ a2+ a3+ . . .converge. As before, we look at the sequence of the absolute values of the ratios of consecutiveterms |an+1|/|an| (still sequence, not series!). If this sequence has a limit L (which must necessarilybe non-negative) andL ≡ limn−→∞|an+1||an|< 1, thenXanconverges. (5)On the other hand, ifL ≡ limn−→∞|an+1||an|> 1 or|an+1||an|−→ ∞, thenXandiverges. (6)3Finally, ifL ≡ limn−→∞|an+1||an|= 1, then this test says nothing. (7)In this last case, you’ll need to find some other way to determine if the seriesPanconverges.For example, for the seriesX(−1)n7nn!, an= (−1)n7n/n! and thuslimn−→∞|an+1||an|= limn−→∞7n + 1=7∞ + 1= 0.Since 0 < 1, this series converges. We will