Strategy for Testing Series We now have several ways of testing a series for convergence or divergence; the problem is to decide which test to use on which series. In this respect testing series is similar to inte- grating functions. Again there are no hard and fast rules about which test to apply to a given series, but you may find the following advice of some use. It is not wise to apply a list of the tests in a specific order until one finally works. That would be a waste of time and effort. Instead, as with integration, the main strategy is to classify the series according to its form. 1. If the series is of the form ͸ 1͞n p, it is a p-series, which we know to be conver- gent if p Ͼ 1 and divergent if p ഛ 1. Ϫ 2. If the series has the form ͸ ar n 1 or ͸ ar n, it is a geometric series, which con- verges if Խ r Խ Ͻ 1 and diverges if Խ r Խ ജ 1. Some preliminary algebraic manipula- tion may be required to bring the series into this form. 3. If the series has a form that is similar to a p-series or a geometric series, then one of the comparison tests should be considered. In particular, if an is a rational function or algebraic function of n (involving roots of polynomials), then the series should be compared with a p-series. (The value of p should be chosen as in Section 8.3 by keeping only the highest powers of n in the numerator and denominator.) The comparison tests apply only to series with positive terms, but if ͸ a has some negative terms, then we can apply the Comparison Test to ͸ n Խ an Խ and test for absolute convergence. 4. If you can see at a glance that lim n l ϱ an 0, then the Test for Divergence should be used. ͸ nϪ1 ͸ n 5. If the series is of the form ͑Ϫ1͒ bn or ͑Ϫ1͒ bn, then the Alternating Series Test is an obvious possibility. 6. Series that involve factorials or other products (including a constant raised to the nth power) are often conveniently tested using the Ratio Test. Bear in mind that Խ anϩ1͞an Խl 1 as n l ϱ for all p-series and therefore all rational or algebraic functions of n. Thus, the Ratio Test should not be used for such series. ෇ ͑ ͒ xϱ ͑ ͒ 7. If an f n , where 1 f x dx is easily evaluated, then the Integral Test is effec- tive (assuming the hypotheses of this test are satisfied). In the following examples we don’t work out all the details but simply indicate which tests should be used. ϱ n Ϫ 1 EXAMPLE 1 ͚ n෇1 2n ϩ 1 l 1 l ϱ Since an 2 0 as n , we should use the Test for Divergence. ϱ sn 3 ϩ 1 EXAMPLE 2 ͚ 3 2 n෇1 3n ϩ 4n ϩ 2 Since a is an algebraic function of n, we compare the given series with a p-series. n ͸ The comparison series for the Limit Comparison Test is bn, where sn 3 n 3͞2 1 b ෇ ෇ ෇ n 3n 3 3n 3 3n 3͞2 1 2 ■ STRATEGY FOR TESTING SERIES ϱ Ϫ 2 EXAMPLE 3 ͚ ne n n෇1 xϱ Ϫx2 Since the integral 1 xe dx is easily evaluated, we use the Integral Test. The Ratio Test also works. ϱ n 3 ͑Ϫ ͒n EXAMPLE 4 ͚ 1 4 n෇1 n ϩ 1 Since the series is alternating, we use the Alternating Series Test. ϱ 2k EXAMPLE 5 ͚ k෇1 k! Since the series involves k!, we use the Ratio Test. ϱ 1 EXAMPLE 6 ͚ n n෇1 2 ϩ 3 Since the series is closely related to the geometric series ͸ 1͞3n, we use the Comparison Test. Exercises ϱ ϱ ͑Ϫ ͒nϪ1 ͞ 1 A Click here for answers. S Click here for solutions. 17. ͚ ͑Ϫ1͒n21 n 18. ͚ n෇1 n෇2 sn Ϫ 1 ϱ ϱ 1–34 Test the series for convergence or divergence. ln n k ϩ 5 19. ͚ ͑Ϫ1͒n 20. ͚ ϱ ϱ s k n 2 Ϫ 1 n Ϫ 1 n෇1 n k෇1 5 1. ͚ 2 2. ͚ 2 ϱ 2n ϱ 2 ෇ n ϩ n ෇ n ϩ n ͑Ϫ2͒ sn Ϫ 1 n 1 n 1 21. ͚ 22. ͚ n 3 ϩ 2 ϩ ϱ ϱ Ϫ n෇1 n n෇1 n 2n 5 1 nϪ1 n 1 3. ͚ 4. ͚ ͑Ϫ1͒ ϱ ϱ 2 ϩ 2 ϩ cos͑n͞2͒ n෇1 n n n෇1 n n ͑ ͞ ͒ 23. ͚ tan 1 n 24. ͚ 2 ෇ ෇ n ϩ 4n ϱ ͑Ϫ3͒nϩ1 ϱ 1 n 1 n 1 ϱ ϱ 2 5. ͚ 3n 6. ͚ 2 n! n ϩ 1 n෇1 2 n෇1 n ϩ n cos n 25. ͚ n2 26. ͚ n n෇1 e n෇1 5 ϱ 1 ϱ 2 k k! 7. ͚ 8. ͚ ϱ k ln k ϱ e 1͞n n෇2 nsln n k෇1 ͑k ϩ 2͒! 27. ͚ 3 28. ͚ 2 k෇1 ͑k ϩ 1͒ n෇1 n ϱ ϱ 2 Ϫk 2 Ϫn3 ϱ Ϫ1 ϱ s 9. ͚ k e 10. ͚ n e tan n j j k෇1 n෇1 29. ͚ 30. ͚ ͑Ϫ1͒ n෇1 nsn j෇1 j ϩ 5 ϱ nϩ1 ϱ ͑Ϫ1͒ n ϱ k ϱ ͑Ϫ ͒n 5 1 11. ͚ 12. ͚ 1 2 n෇2 n ln n n෇1 n ϩ 25 31. ͚ k k 32. ͚ ln n k෇1 3 ϩ 4 n෇2 ͑ln n͒ ϱ n 2 ϱ 3 n ϱ sin͑1͞n͒ ϱ 13. ͚ 14. ͚ sin n 33. ͚ 34. ͚ (sn 2 Ϫ 1) n෇1 n! n෇1 n෇1 sn n෇1 ϱ n! 15. ͚ n෇0 2 ؒ 5 ؒ 8 ؒ иии ؒ ͑3n ϩ 2͒ ϱ n 2 ϩ 1 16. ͚ 3 n෇1 n ϩ 1 STRATEGY FOR TESTING SERIES ■ 3 Answers S Click here for solutions. 1. D 3. C 5. C 7. D 9. C 11. C 13. C 15. C 17. D 19. C 21. C 23. D 25. C 27. C 29. C 31. D 33. C 4 ■ STRATEGY FOR TESTING SERIES Solutions: Strategy for Testing Series n2 1 1 1/n2 ∞ n2 1 1. lim an =lim − =lim − =1=0,sotheseries − diverges by the Test for n n n2 +1 n 1+1/n 6 n2 +1 →∞ →∞ →∞ n=1 X Divergence. 1 1 ∞ 1 ∞ 1 3. 2 < 2 for all n 1,so 2 converges by the Comparison Test with 2 ,ap-series that n + n n ≥ n=1 n + n n=1 n converges because p =2> 1. P P n+2 3n 3n an+1 ( 3) 2 3 2 3 3 5. lim =lim −3(n+1) n+1 =lim −3n· 3 =lim 3 = < 1,sotheseries n an n 2 · ( 3) n 2 2 n 2 8 →∞ ¯ ¯ →∞ ¯ − ¯ →∞ ¯ · ¯ →∞ ¯ n+1¯ ¯ ¯ ¯ ¯ ∞ (¯ 3) ¯ ¯ ¯ ¯ ¯ −¯ ¯ is absolutely¯ convergent by the¯ Ratio Test.¯ ¯ 23n n=1 X 1 7. Let f(x)= .Thenf is positive, continuous, and decreasing on [2, ), so we can apply the Integral Test. x √ln x ∞ 1 u =lnx, 1/2 1/2 Since dx = u− du =2u + C =2√ln x + C,wefind x√ln x du = dx/x Z " # Z t ∞ dx dx t =lim =lim 2√ln x =lim 2 √ln t 2 √ln 2 = .Sincetheintegral 2 x √ln x t 2 x √ln x t 2 t − ∞ Z →∞ Z →∞ h i →∞ ³ ´ ∞ 1 diverges, the given series diverges. √ n=2 n ln n X 2 ∞ 2 k ∞ k 9. k e− = . Using the Ratio Test, we get ek k=1 k=1 X X 2 k 2 ak+1 (k +1) e k +1 1 2 1 1 lim =lim k+1 2 =lim =1 = < 1, so the series converges. k ak k e · k k k · e · e e →∞ ¯ ¯ →∞ ¯ ¯ →∞ "µ ¶ # ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ n+1 ¯ 1 ¯ ¯ ¯ ∞ ( 1) 11. bn = > 0 for n 2, bn is decreasing, and lim bn =0,sothegivenseries − converges by n ln n n n ln n ≥ { } →∞ n=2 the Alternating Series Test. P n+1 2 2 an+1 3 (n +1) n! 3(n +1) n +1 13. lim =lim n 2 =lim 2 =3 lim 2 =0< 1,sotheseries n an n (n +1)! · 3 n n (n +1)n n n →∞ ¯ ¯ →∞ ¯ ¯ →∞ · ¸ →∞ ¯n 2 ¯ ¯ ¯ ∞ 3¯ n ¯ ¯ ¯ ¯ converges¯ by¯ the Ratio Test. ¯ n! n=1 X an+1 (n +1)! 2 5 8 (3n +2) 15. lim =lim · · ····· n an n 2 5 8 (3n +2)[3(n +1)+2] · n! →∞ ¯ ¯ →∞ ¯ · · ····· ¯ ¯ ¯ ¯n +1 1 ¯ ¯ ¯=lim¯ = < 1 ¯ ¯ ¯ n ¯3n +5 3 ¯ →∞ ∞ n! so the series converges by the Ratio Test. 2 5 8 (3n +2) n=0 X · · ····· ∞ 17. lim 21/n =20 =1,so lim ( 1)n 21/n does not exist and the series ( 1)n21/n diverges by the n n →∞ →∞ − n=1 − X Test for Divergence. STRATEGY FOR TESTING SERIES ■ 5 ln x 2 ln x 2 ln n 2 19. Let f(x)= .Thenf 0(x)= − < 0 when ln x>2 or x>e ,so is decreasing for n>e . √x 2x3/2 √n ln n 1/n 2 ∞ ln n By l’Hospital’s Rule, lim = lim = lim =0,sotheseries ( 1)n converges by n √n n 1/ (2√n) n √n √n →∞ →∞ →∞ n=1 − X the Alternating Series Test.