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Section 11.4
- 1. Chapter 11: Sequences,Series, and Power Series Section 11.4: Comparison Tests Alea Wittig SUNY Albany
- 2. Outline The Direct ComparisonTest The Limit Comparison Test
- 3. ▶ Consider theseries ∞ X n=1 1 2n + 1 ▶ This series looks very much like a convergent geometric series we have examined: ∞ X n=1 1 2n = 1 ▶ But P 1 2n+1 isn’t quite a geometric series, since it can’t be written in the form P arn−1 ▶ Nevertheless, we can use what we know about the series P 1 2n to determine whether the series P 1 2n+1 converges as well.
- 4. ▶ Let sndenote the nth partial sum of P 1 2n+1: sn = n X i=1 1 2i + 1 ▶ Observe that 0 < 1 2i + 1 < 1 2i for all integers i ▶ So adding up the first n terms we have 0 < n X i=1 1 2i + 1 | {z } sn < n X i=1 1 2i ≤ 1 since n X i=1 1 2i < ∞ X n=1 1 2n = 1 therefore 0 < sn < 1 for all n, ie, sn is bounded
- 5. ▶ Since 1 2n+1+1 ispositive for any n, n X i=1 1 2i + 1 < n X i=1 1 2i + 1 + 1 2n+1 + 1 sn < sn+1, ie, sn is increasing ▶ So by the bounded and monotonic convergence theorem, sn = Pn i=1 1 2i +1 is a bounded, increasing sequence and thus it is convergent. ▶ Thus by definition of a convergent series, ∞ X n=1 1 2n + 1 is convergent
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- 7. The Direct ComparisonTest Theorem: Suppose that ∞ X n=1 an and ∞ X n=1 bn are series with positive terms. (i) If P bn is convergent and an ≤ bn for all n, then P an is also convergent. (ii) If P bn is divergent and an ≥ bn for all n, then P an is also divergent.
- 8. On the leftis an illustration of Direct Comparison (i). On the right is an illustration of Direct Comparison (ii). Here tn = Pn i=1 bi , the partial sum of bn. And sn = Pn i=1 ai , the partial sum of an.
- 9. ▶ In general,to use direct comparison on the series P an, we need a known series P bn to compare it to. ▶ P bn will most often be: ▶ A geometric series: X arn−1 ▶ converges if |r| < 1 and diverges if |r| ≥ 1. ▶ A p-series: X 1 np ▶ converges if p > 1 and diverges if p ≤ 1.
- 10. Example 1 1/3 Determine whetherthe series ∞ X n=1 5 2n2 + 4n + 3 converges or diverges.
- 11. Example 1 2/3 ▶ Touse direct comparison we first check that we indeeed have positive terms ✓ ▶ Next we must find a series P bn to compare with. ∞ X n=1 an = ∞ X n=1 5 2n2 + 4n + 3 looks like a p-series so we find a p-series to compare to. ▶ As a general strategy for picking bn, take the numerator of an over the highest power term in the denominator of an: ∞ X n=1 bn = ∞ X n=1 5 2n2
- 12. Example 1 3/3 ▶ Forall n we have 5 2n2 + 4n + 3 < 5 2n2 ie, an < bn. ▶ So if the series P bn converges, then direct comparison will tell us that P an converges as well. ∞ X n=1 5 2n2 = 5 2 ∞ X n=1 1 n2 converges p = 2 > 1 =⇒ ∞ X n=1 5 2n2 + 4n + 3 converges by direct comparison
- 13. Example 2 1/2 Test theseries ∞ X k=1 ln k k for convergence or divergence.
- 14. Example 2 2/2 ▶ Herewe have a series that doesn’t look quite like a p-series or a geometric series. ▶ Nevertheless, we know that ln k > 1 for k > e, and the harmonic series ∞ X k=1 1 k = 1 + 1 2 + ∞ X k=3 1 k is divergent ▶ So since ln k k > 1 k for k ≥ 3 by direct comparison, ∞ X k=3 ln k k is divergent and thus ∞ X k=1 ln k k = ln 2 2 + ∞ X k=3 ln k k is divergent
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- 16. ▶ To determinewhether the series ∞ X n=1 1 2n − 1 converges or diverges, then we might think to use direct comparison test since the terms 1 2n−1 are close to 1 2n . ▶ But this won’t work since, for all n, 1 2n − 1 > 1 2n and X 1 2n converges. ▶ For this example we can use the following test
- 17. The Limit ComparisonTest Theorem: Suppose that X an and X bn are series with positive terms. If lim n→∞ an bn = c where c > 0 is a finite number then either both series converge or both diverge.
- 18. Example 3 1/2 Use theLimit Comparison Test to test ∞ X n=1 1 2n − 1 for convergence or divergence.
- 19. Example 3 2/2 ▶ Wecompare 1 2n − 1 with 1 2n , which we know converges. lim n→∞ an bn = lim n→∞ 1 2n 1 2n − 1 = lim n→∞ 2n − 1 2n = lim n→∞ 1 − 1 2n = 1 ▶ So the Limit Comparison Test tells us that ∞ X n=1 1 2n − 1 converges.
- 20. Example 4 1/3 Determine whetherthe sequence ∞ X n=1 2n2 + 3n √ 5 + n5 converges or diverges.
- 21. Example 4 2/3 ▶ Theseries has positive terms ✓ ▶ Using our general strategy for choosing P bn, we compare to the following series: 2n2 + 3n √ n5 ∞ X n=1 2n2 + 3n √ n5 = ∞ X n=1 2n2 √ n5 + 3n √ n5 = ∞ X n=1 2 n1/2 + 3 n3/2 > ∞ X n=1 2 n1/2 ← diverges ▶ We can’t use direct comparison because 2n2 + 3n √ n5 > 2n2 + 3n √ 5 + n5
- 22. Example 4 3/3 ▶ Buttaking the limit of the ratio of the terms of the series: lim n→∞ 2n2 + 3n √ 5 + n5 2n2 + 3n √ n5 = lim n→∞ 2n2 + 3n √ 5 + n5 √ n5 2n2 + 3n = lim n→∞ √ n5 √ 5 + n5 = s lim n→∞ n5 5 + n5 = s lim n→∞ 1 5 n5 + 1 = 1 0 ▶ Thus by Limit Comparison, the series converges.