10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1 10.5 POSITIVE TERM SERIES: COMPARISON TESTS This section discusses how to determine whether some series converge or diverge by comparing them with other series which we already know converge or diverge. In the basic Comparison Test we compare the two series term by term. In the more powerful Limit Comparison Test, we compare limits of ratios of the terms of the two series. Finally, we focus on the parts of the terms of a series that determine whether the series converges or diverges. Comparison Test Informally, if the individual terms of our series are smaller than the terms of a known convergent series, then our series converges. If our series is larger, term by term, than a known divergent series then our series diverges. If the individual terms of our series are larger than the terms of a convergent series or smaller than the terms of a divergent series, then our series may converge or diverge –– the Comparison Test does not tell us. Comparison Test Suppose we want to determine whether ∑k=1∞ ak converges or diverges. (a) If there is a convergent series ∑k=1∞ ck with 0 < ak ≤ ck for all k, then ∑k=1∞ ak converges. (b) If there is a divergent series ∑k=1∞ dk with ak ≥ dk > 0 for all k, then ∑k=1∞ ak diverges. Proof: Since all of the terms of the ak, ck, and dk series are positive, their sequences of partial sums are all monotonic increasing. The proof compares the partial sums of the various series. (a) Suppose that 0 < ak ≤ ck for all k and that ∑k=1∞ ck converges. Since ∑k=1∞ ck converges, then the partial sums tn = ∑k=1n ck approach a finite limit: ! limn"#tn = L. For each n, sn ≤ tn (why?) so ! limn"#sn ≤ ! limn"#tn = L , and the sequence { sn } is10.5 Positive Term Series: Comparison Tests Contemporary Calculus 2 bounded by L. Finally, by the Monotone Convergence Theorem, we can conclude that { sn } converges and that ∑k=1n ak converges. (b) Suppose that ak ≥ dk > 0 for all k and that ∑k=1∞ dk diverges. Since ∑k=1∞ dk diverges, then the partial sums un = ∑k=1n dk approach infinity: ! limn"# un = ∞. Then ! limn"# sn ≥ ! limn"# un = ∞ so the sequence of partial sums { sn } diverges and the series ∑k=1n ak diverges. The Comparison Test requires that we select and compare our series against a series whose convergence or divergence is known, and that choice requires that we know a collection of series that converge and some that diverge. Typically, we pick a p–series or a geometric series to compare with our series, but this choice requires some experience and practice. Example 1: Use the Comparison Test to determine the convergence or divergence of (a) ! 1k2+ 3k=1"# and (b) ∑k=1∞ k + 1k2 . Solution: For these two series it is useful to compare with p–series for appropriate values of p. (a) For all k, 1k2 + 3 < 1k2 , and ∑k=1∞ 1k2 converges (P–Test, p = 2) so ! 1k2+ 3k=1"# converges. (b) For all k, k + 1k2 = 1k + 1k2 > 1k . Since ! 1kk=1"# diverges (P–Test, p = 1), we conclude that ∑k=1∞ k + 1k2 diverges. Practice 1: Use the Comparison Test to determine the convergence or divergence of (a) ! 1k - 2k=3"# and (b) ! 12k+ 7k=1"# .10.5 Positive Term Series: Comparison Tests Contemporary Calculus 3 Example 2: A student has shown algebraically that 1k2 < 1k2 – 1 < 1k for all k ≥ 2. From this information and the Comparison Test, what can the student conclude about the convergence of the series ! 1k2"1k=2#$ ? Solution: Nothing. The Comparison Test only gives a definitive answer if our series is smaller than a convergent series or if our series is larger than a divergent series. In this example, our series is larger than a convergent series, ∑ 1k2 , and is smaller than a divergent series, ∑ 1k , so we can not conclude anything about the convergence of ! 1k2"1k=2#$. However, we can show that if k ≥ 2 then 1k2 – 1 < 2k2 . Since ! 2k2k=2"# converges (P–Test), we can conclude that ! 1k2"1k=2#$ converges. Next in this section we present a variation on the Comparison Test that allows us to quickly conclude that ! 1k2"1k=2#$ converges. Limit Comparison Test The exact value of the sum of a series depends on every part of the terms of the series, but if we are only asking about convergence or divergence, some parts of the terms can be safely ignored. For example, the three series with terms 1/k2 , 1/(k2 + 1) , and 1/(k2 – 1) converge to different values, ! 1k2k=2"# ≈ 0.645 , ! 1k2+ 1k=2"# ≈ 0.577 , ! 1k2"1k=2#$ = 0.750 , but they all do converge. The "+ 1" and "– 1" in the denominators effect the value of the final sum, but they do not effect whether that sum is finite or infinite. When k is a large number, the values of 1/(k2 + 1) and 1/(k2 – 1) are both very close to the value of 1/k2 , and the convergence or divergence of the series ! 1k2+ 1k=2"# and ! 1k2"1k=2#$ can be predicted from the convergence or divergence of the series ∑k=2∞ 1k2 . The Limit Comparison Test states these ideas precisely.10.5 Positive Term Series: Comparison Tests Contemporary Calculus 4 Limit Comparison Test Suppose ak > 0 for all k, and we want to determine whether ∑k=1∞ ak converges or diverges. If there is a series ∑k=1∞ bk so that limk∅∞ akbk = L, a positive, finite value, then ∑k=1∞ ak and ∑k=1∞ bk both converge or both diverge. Idea for a proof: The key idea is that if ! limk"# akbk = L is a positive, finite value, then, when n is very large, akbk ≈ L so ak ≈ L.bk and ∑k=N∞ ak ≈ L. ∑k=N∞ bk . If one of these series converges, then so does the other. If one of these series diverges, then so does the other. When n is a relatively small number, the ak and bk values may not have a ratio close to L, but the first "few" values of a series do not effect the convergence or divergence of the series. A proof of the Limit Comparison Test is given in an Appendix after the Practice Answers. Example 3: Put ak = 1k2 , bk = 1k2 + 1 , and ck = 1k2 – 1 and show that ! limk"# akbk = 1 and ! limk"# akck = 1. Since ∑k=2∞ 1k2 converges (P–Test, p =