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Section 11.7
- 1. Chapter 11: Sequences, Series, and Power Series Section 11.7: Strategy for Testing Series Alea Wittig SUNY Albany
- 3. Strategies
- 4. Test for Divergence lim n→∞ an ̸= 0 =⇒ ∞ X n=1 an diverges ▶ The first test you might consider doing is the test for divergence, as it is the simplest. ▶ If lim n→∞ an = 0 then test for divergence tells us nothing at all so be careful not to make that mistake.
- 5. Test for Divergence eg ∞ X n=1 5n3 + n2 + 1 n2 + 3n ; Rational with higher degree in numerator =⇒ the numerator dominates so lim n→∞ an = ∞ eg ∞ X n=1 5n2 + n2 + 1 n2 + 2n + 1 ; Rational with same degree in numerator and denominator =⇒ an approaches ratio of leading coefficients lim n→∞ an = 5 eg ∞ X n=1 (−1)n ; an oscillates between two values, never getting closer to either lim n→∞ an DNE
- 6. p-Series ▶ p-Series X 1 np converges iff p > 1 ▶ If you see a p−series, use the p-series test.
- 7. Geometric Series If we can write the series in the form ∞ X n=1 arn−1 then the series converges iff |r| < 1. ▶ If it converges, then it converges to a 1 − r ▶ Sometimes we need to manipulate the expression an in order to get this form. eg ∞ X n=1 5n+1 7−n = ∞ X n=1 5 5 7 n = ∞ X n=1 25 7 5 7 n−1 = 25 2
- 8. Comparison Tests ▶ Use comparison Test: ▶ If P an is similar to a p-series. ▶ If an is a rational function, or an algebraic function (involving roots of polynomials), compare to a p-series. Chose bn by keeping the numerator of an and just the highest power in the denominator of an. eg To test P n5 +3n2 n7+4 , compare to P n5 +3n2 n7 = P 1 n2 + 3 n5 ▶ If P an is similar to a geometric series. ▶ Proceed similarly, eg To test P 5n 7n+4n , compare to the geo series P 5n 7n
- 9. Comparison Tests ▶ Even though the comparison test is only for series with positive terms, we can still use it on P |an| to test for absolute convergence. If it is absolutely convergent, then it is convergent. If it is not absolutely convergent, it could still be convergent so consider a different test. ▶ When doing comparison: ▶ First determine series P bn what to compare to. a) Does P bn converge where an ≤ bn for each n? b) Does P bn diverge where bn ≤ an for each n? ▶ If either a) or b) is true, direct comparison works. ▶ If neither is true then do limit comparison.
- 10. Example 1 1/2 Determine whether the following series converges ∞ X n=1 1 √ n2 + 1
- 11. Example 1 2/2 ▶ Test for divergence won’t work since lim n→∞ 1 √ n2 + 1 = 0. ▶ Looks like a p-Series, so use a comparison test. ▶ Highest power term in denominator is √ n2 so compare to P 1 √ n2 = P 1 n , which diverges. ▶ Since n2 + 1 ≥ n2 =⇒ p n2 + 1 ≥ √ n2 = n =⇒ 1 √ n2 + 1 ≤ 1 n , we can’t use direct comparison. Try limit comparison: lim n→∞ 1 √ n2+1 1 n = lim n→∞ n √ n2 + 1 = 1 So both series diverge.
- 12. Alternating Series Test ▶ Use AST whenever you have a series of the form X (−1)n−1 bn or X (−1)n bn bn 0 ▶ Note: If ∞ X n=1 bn converges then the given series is absolutely convergent and therefore convergent.
- 13. Ratio Test ▶ Use ratio test if ▶ P an involves factorials. ▶ P an involves multiple products including constant to power n. ▶ Don’t use for p−series (or any rational or algebraic functions on n), you will always get L = 1.
- 14. Root Test ▶ Use root test if ▶ an is of the form (bn)n ▶ If root test is inconclusive then ratio test will also be inconclusive.
- 15. Integral Test ▶ Use integral test if ▶ an = f (n) where R ∞ 1 f (x)dx is easily computed and f is positive, continuous, and decreasing (or eventually decreasing.)
- 16. Examples
- 17. Example 2 1/2 Determine which test or tests to use for convergence of the following series. Then test it. ∞ X n=1 n − 1 2n + 1
- 18. Example 2 2/2 ▶ n−1 2n+1 is improper rational and lim n→∞ n − 1 2n + 1 = 1 2 so the series diverges by the test for divergence.
- 19. Example 3 1/3 Determine which test or tests to use for convergence of the following series. Then test it. ∞ X n=1 ne−n2
- 20. Example 3 2/3 ▶ an = f (n) where f (x) = x ex2 . ▶ f is clearly positive and continuous for x 0. ▶ f is decreasing: f ′ (x) = ex2 − x(2x)ex2 e2x2 = ex2 (1 − 2x2) e2x2 = 1 − 2x2 ex2 0 ⇐⇒ 1 − 2x2 0 ⇐⇒ 1 √ 2 x ✓ ▶ Let’s use the integral test.
- 21. Example 3 3/3 ▶ Now computing the improper integral we use a simple u-substitution: Z ∞ 1 xe−x2 dx = lim t→∞ Z t 1 xe−x2 dx = lim t→∞ − 1 2 Z −t2 −1 eu du [u = −x2 du = −2xdx] = − 1 2 lim t→∞ 1 et2 − 1 e = 1 2e ▶ Since the integral converges, the series P ne−n2 converges as well.
- 22. Example 4 1/2 Determine which test or tests to use for convergence of the following series. Then test it. ∞ X k=1 2k k!
- 23. Example 4 2/2 ▶ an = 2k k! has a constant raised to n as well as a factorial, so let’s use ratio test. lim k→∞
- 26. ak+1 ak
- 29. = lim k→∞ 2k+1 (k+1)! 2k k! = lim k→∞ 2k+1 (k + 1)! k! 2k = lim k→∞ 2k+1 2k k! (k + 1)! = lim k→∞ 2 · 2k 2k k! (k + 1) k! = lim k→∞ 2 k + 1 = 0 1 ▶ Therefore by the ratio test, the series converges.
- 30. Example 5 1/2 Determine which test or tests to use for convergence of the following series. Then test it. ∞ X n=1 sin 2n 1 + 2n
- 31. Example 5 2/2 ▶ Test for absolute convergence and thus convergence using direct comparison: 0 ≤ | sin 2n| ≤ 1 =⇒ 0 ≤
- 34. sin 2n 1 + 2n
- 37. ≤ 1 1 + 2n and since P 1 1+2n is similar to the geo series P 1 2n , we use a comparison test. ▶ P 1 2n converges and 1 1 + 2n ≤ 1 2n therefore P 1 1+2n converges by direct comparison. ▶ Now using direct comparison again, P |an| = P
- 39. sin 2n 1+2n
- 41. converges. ▶ Therefore P an = P sin 2n 1+2n is absolutely convergent and thus convergent.