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 1 n p , it is a p -series, which we know to be conver- gent if p 1 and divergent if p 1 . 2. If the series has the form ar n 1 or ar n , it is a geometric series, which con- verges if r 1 and diverges if r 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 an and test for absolute convergence.
4. If you can see at a glance that lim n l an 0 , then the Test for Divergence should be used. n 1 n 5. If the series is of the form 1 bn or 1 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
an 1 an l 1 as n l for all p -series and therefore all rational or algebraic functions of n . Thus, the Ratio Test should not be used for such series. x 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.
n 1 EXAMPLE 1 n 1 2n 1 l 1 l Since an 2 0 as n , we should use the Test for Divergence.
sn 3 1 EXAMPLE 2 3 2 n 1 3n 4n 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 3 2 1 b n 3n 3 3n 3 3n 3 2
1 2 ■ STRATEGY FOR TESTING SERIES
2 EXAMPLE 3 ne n n 1 x x2 Since the integral 1 xe dx is easily evaluated, we use the Integral Test. The Ratio Test also works.
n 3 n EXAMPLE 4 1 4 n 1 n 1 Since the series is alternating, we use the Alternating Series Test.
2k EXAMPLE 5 k 1 k! Since the series involves k! , we use the Ratio Test.
1 EXAMPLE 6 n n 1 2 3 Since the series is closely related to the geometric series 1 3n , we use the Comparison Test.
Exercises
n 1 1 A Click here for answers. S Click here for solutions. 17. 1 n21 n 18. n 1 n 2 sn 1 1–34 Test the series for convergence or divergence. ln n k 5 19. 1 n 20. s k n 2 1 n 1 n 1 n k 1 5 1. 2 2. 2 2n 2 n n n n 2 sn 1 n 1 n 1 21. 22. n 3 2 n 1 n n 1 n 2n 5 1 n 1 n 1 3. 4. 1 2 2 cos n 2 n 1 n n n 1 n n 23. tan 1 n 24. 2 n 4n 3 n 1 1 n 1 n 1 2 5. 3n 6. 2 n! n 1 n 1 2 n 1 n n cos n 25. n2 26. n n 1 e n 1 5 1 2 k k! 7. 8. k ln k e 1 n n 2 nsln n k 1 k 2 ! 27. 3 28. 2 k 1 k 1 n 1 n 2 k 2 n3 1 s 9. k e 10. n e tan n j j k 1 n 1 29. 30. 1 n 1 nsn j 1 j 5 n 1 1 n k n 5 1 11. 12. 1 2 n 2 n ln n n 1 n 25 31. k k 32. ln n k 1 3 4 n 2 ln n n 2 3 n sin 1 n 13. 14. sin n 33. 34. (sn 2 1) n 1 n! n 1 n 1 sn n 1 n! 15. n 0 2 5 8 3n 2 n 2 1 16. 3 n 1 n 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.
2n n ∞ ( 2) ∞ 4 n 4 21. − = . lim an =lim =0< 1, so the given series is absolutely convergent by the nn n n | | n n n=1 n=1µ ¶ →∞ →∞ RootP Test. P p 1 1 23. Using the Limit Comparison Test with a =tan and b = ,wehave n n n n µ ¶ 2 2 an tan(1/n) tan(1/x) H sec (1/x) ( 1/x ) 2 2 lim =lim = lim = lim · 2− =limsec (1/x)=1 =1> 0. n bn n 1/n x 1/x x 1/x x →∞ →∞ →∞ →∞ − →∞ Since n∞=1 bn is the divergent harmonic series, n∞=1 an is also divergent.
n2 n2 P an+1 (n +1)!P e (n +1)n! e n +1 25. Use the Ratio Test. lim =lim 2 =lim 2 · =lim 2n+1 =0< 1,so n an n ¯ e(n+1) · n! ¯ n en +2n+1n! n e →∞ ¯ ¯ →∞ ¯ ¯ →∞ →∞ ¯ ¯ ¯ ¯ ∞ n! ¯ ¯ ¯ ¯ converges. ¯ ¯ ¯ ¯ en2 n=1 X t ∞ ln x ln x 1 H ln n 27. dx =lim (using integration by parts) =1.So∞ converges by the Integral Test, x2 t − x − x n2 Z2 →∞ · ¸1 n=1 k ln k k ln k ln k ∞ k ln k P and since 3 < 3 = 2 , the given series 3 converges by the Comparison Test. (k +1) k k k=1 (k +1)
1 P tan− n π/2 ∞ π/2 π ∞ 1 3 29. 0 < 3/2 < 3/2 . 3/2 = 3/2 which is a convergent p-series (p = 2 > 1), so n n n=1 n 2 n=1 n 1 P P ∞ tan− n 3/2 converges by the Comparison Test. n=1 n
P k k k k 5 k (5/4) 3 5 31. lim ak = lim =[divide by 4 ]lim = since lim =0and lim = . k k 3k +4k k (3/4)k +1 ∞ k 4 k 4 ∞ →∞ →∞ →∞ →∞µ ¶ →∞µ ¶ ∞ 5k Thus, diverges by the Test for Divergence. 3k +4k k=1 X sin(1/n) 1 an sin(1/n) ∞ sin(1/n) 33. Let an = and bn = .Then lim =lim =1> 0,so converges by √n n√n n bn n 1/n √n →∞ →∞ n=1 X ∞ 1 limit comparison with the convergent p-series (p =3/2 > 1). n3/2 n=1 X