Section 11.8 Power Series

(1)

Section 11.8 Power Series

36. Find the radius of convergence and interval of convergence of the series.

∞

P

n=1

n!xⁿ 1·3·5···(2n−1). Solution:

SECTION 11.8 POWER SERIES ¤ 165 24.  = ²^

2 · 4 · 6 · · · (2) = ²^

2^! = ^ 2^( − 1)!, so

lim→∞



+1





 = lim→∞

( + 1) ||^+1

2^+1! ·2^( − 1)!

 ||^ = lim

→∞

 + 1

²

||

2 = 0. Thus, by the Ratio Test, the series converges for all real  and we have  = ∞ and  = (−∞ ∞).

25. If = (5 − 4)^

³ , then

lim→∞



+1





 = lim→∞



(5 − 4)^+1

( + 1)³ · ³ (5 − 4)^



 = lim→∞|5 − 4|

 

 + 1

3

= lim

→∞|5 − 4|

 1

1 + 1

3

= |5 − 4| · 1 = |5 − 4|

By the Ratio Test, ^∞

=1

(5 − 4)^

³ converges when |5 − 4|  1 ⇔ 

 − ⁴5

  ¹₅ ⇔ −¹5   − ⁴5  ¹₅ ⇔

3

5    1, so  = ¹₅. When  = 1, the series ^∞

=1

1

³ is a convergent -series (  = 3  1). When  = ³₅, the series

∞

=1

(−1)^

³ converges by the Alternating Series Test. Thus, the interval of convergence is  =3 5 1.

26. If = ^2

 (ln )², then lim

→∞



+1





 = lim→∞



 ^2+2

( + 1)[ln( + 1)]² · (ln )²

^2



 =² lim

→∞

 (ln )²

( + 1)[ln( + 1)]² = ². By the Ratio Test, the series^∞

=2

^2

 (ln )² converges when ² 1 ⇔ ||  1, so  = 1. When  = ±1, ^2= 1, the series ^∞

=2

1

 (ln )² converges by the Integral Test (see Exercise 11.3.29). Thus, the interval of convergence is  = [−1 1].

27. If  = ^

1 · 3 · 5 · · · (2 − 1), then

lim→∞



+1





 = lim→∞



 ^+1

1 · 3 · 5 · · · (2 − 1)(2 + 1) ·1 · 3 · 5 · · · (2 − 1)

^



 = lim→∞

||

2 + 1 = 0  1. Thus, by the Ratio Test, the series^∞

=1

^

1 · 3 · 5 · · · (2 − 1) converges for all real  and we have  = ∞ and  = (−∞ ∞).

28. If  = ! ^

1 · 3 · 5 · · · (2 − 1), then

→∞lim



+1





 = lim→∞



 ( + 1)! ^+1

1 · 3 · 5 · · · (2 − 1)(2 + 1) ·1 · 3 · 5 · · · (2 − 1)

! ^



 = lim→∞

( + 1) ||

2 + 1 = ¹₂||.

By the Ratio Test, the series^∞

=1

converges when ¹₂||  1 ⇒ ||  2 so  = 2. When  = ±2,

|^| = ! 2^

1 · 3 · 5 · · · (2 − 1) = [1 · 2 · 3 · · · ] 2^

[1 · 3 · 5 · · · (2 − 1)] = 2 · 4 · 6 · · · 2

1 · 3 · 5 · · · (2 − 1)  1, so both endpoint series diverge by the Test for Divergence. Thus, the interval of convergence is  = (−2 2).

29. (a) We are given that the power series∞

=0^is convergent for  = 4. So by Theorem 4, it must converge for at least

−4   ≤ 4. In particular, it converges when  = −2; that is,_∞

=0(−2)^is convergent.

38. Suppose that

∞

P

n=0

cnxⁿ converges when x = −4 and diverges when x = 6. What can be said about the convergence or divergence of the following series?

(a)

∞

P

n=0

c_n (b)

∞

P

n=0

c_n8ⁿ (c)

∞

P

n=0

c_n(−3)ⁿ (d)

∞

P

n=0

(−1)ⁿc_n9ⁿ Solution:

166 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES (b) It does not follow that_∞

=0(−4)^is necessarily convergent. [See the comments after Theorem 4 about convergence at the endpoint of an interval. An example is = (−1)^(4^).]

30. We are given that the power series_∞

=0^is convergent for  = −4 and divergent when  = 6. So by Theorem 4 it converges for at least −4 ≤   4 and diverges for at least  ≥ 6 and   −6. Therefore:

(a) It converges when  = 1; that is,

is convergent.

(b) It diverges when  = 8; that is,

8^is divergent.

(c) It converges when  = −3; that is,

(−3^)is convergent.

(d) It diverges when  = −9; that is,

(−9)^=

(−1)^9^is divergent.

31. If = (!)^

()!^, then

lim→∞



+1





 = lim→∞

[( + 1)!]^()!

(!)^[( + 1)]!|| = lim__→∞ ( + 1)^

( + )( +  − 1) · · · ( + 2)( + 1)||

= lim

→∞

 ( + 1) ( + 1)

( + 1)

( + 2)· · · ( + 1) ( + )



||

= lim

→∞

  + 1

 + 1



lim→∞

  + 1

 + 2



· · · lim__→∞

  + 1

 + 



||

=

1





||  1 ⇔ ||  ^for convergence, and the radius of convergence is  = ^

32. (a) Note that the four intervals in parts (a)–(d) have midpoint  = ¹₂( + )and radius of convergence  = ¹₂( − ). We also know that the power series ^∞

=0

^has interval of convergence (−1 1). To change the radius of convergence to , we can change ^to 





. To shift the midpoint of the interval of convergence, we can replace  with  − . Thus, a power

series whose interval of convergence is ( ) is ^∞

=0

  − 



^

, where  = ¹₂( + )and  = ¹₂( − ).

(b) Similar to Example 2, we know that ^∞

=1

^

 has interval of convergence [−1 1). By introducing the factor (−1)^ in , the interval of convergence changes to (−1 1]. Now change the midpoint and radius as in part (a) to get

∞

=1(−1)^1



  − 



^

as a power series whose interval of convergence is ( ].

(c) As in part (b), ^∞

=1

1



  − 



^

is a power series whose interval of convergence is [ ).

(d) If we increase the exponent on  (to say,  = 2), in the power series in part (c), then when  = , the power series

∞

=1

1

²

  − 



^

will converge by comparison to the p-series with  = 2  1, and the interval of convergence will be [ ].

33. No. If a power series is centered at , its interval of convergence is symmetric about . If a power series has an infinite radius of convergence, then its interval of convergence must be (−∞ ∞), not [0 ∞).

40. Let p and q be real numbers with p < q. Find a power series whose interval of convergence is (a) (p, q) (b) (p, q] (c) [p, q) (d) [p, q]

Solution:

1

(2)

166 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES (b) It does not follow that∞

=0(−4)^is necessarily convergent. [See the comments after Theorem 4 about convergence at the endpoint of an interval. An example is = (−1)^(4^).]

30. We are given that the power series∞

=0^is convergent for  = −4 and divergent when  = 6. So by Theorem 4 it converges for at least −4 ≤   4 and diverges for at least  ≥ 6 and   −6. Therefore:

(a) It converges when  = 1; that is,

is convergent.

(b) It diverges when  = 8; that is,

8^is divergent.

(c) It converges when  = −3; that is,

(−3^)is convergent.

(d) It diverges when  = −9; that is,

(−9)^=

(−1)^9^is divergent.

31. If = (!)^

()!^, then

lim→∞



+1





 = lim→∞

[( + 1)!]^()!

(!)^[( + 1)]!|| = lim__→∞ ( + 1)^

( + )( +  − 1) · · · ( + 2)( + 1)||

= lim

→∞

 ( + 1) ( + 1)

( + 1)

( + 2)· · · ( + 1) ( + )



||

= lim

→∞

  + 1

 + 1



lim→∞

  + 1

 + 2



· · · lim__→∞

  + 1

 + 



||

=

1





||  1 ⇔ ||  ^for convergence, and the radius of convergence is  = ^

32. (a) Note that the four intervals in parts (a)–(d) have midpoint  = ¹₂( + )and radius of convergence  = ¹₂( − ). We also know that the power series ^∞

=0

^has interval of convergence (−1 1). To change the radius of convergence to , we can change ^to 





. To shift the midpoint of the interval of convergence, we can replace  with  − . Thus, a power

series whose interval of convergence is ( ) is ^∞

=0

  − 



^

, where  = ¹₂( + )and  = ¹₂( − ).

(b) Similar to Example 2, we know that ^∞

=1

^

 has interval of convergence [−1 1). By introducing the factor (−1)^ in , the interval of convergence changes to (−1 1]. Now change the midpoint and radius as in part (a) to get

∞

=1(−1)^ 1



  − 



^

as a power series whose interval of convergence is ( ].

(c) As in part (b), ^∞

=1

1



  − 



^

is a power series whose interval of convergence is [ ).

(d) If we increase the exponent on  (to say,  = 2), in the power series in part (c), then when  = , the power series

∞

=1

1

²

  − 



^

will converge by comparison to the p-series with  = 2  1, and the interval of convergence will be [ ].

33. No. If a power series is centered at , its interval of convergence is symmetric about . If a power series has an infinite radius of convergence, then its interval of convergence must be (−∞ ∞), not [0 ∞).

44. Suppose that the power seriesP cn(x − a)ⁿsatisfies cn6= 0 for all n. Show that if lim

n→∞|_c^cⁿ

n+1| exists, then it is equal to the radius of convergence of the power series.

Solution:

168 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 2 · 3 · 5 · 6 · · · (3 − 1) · 3 = (3)!

1 · 4 · 7 · · · (3 − 2) = (3)!



=1(3 − 2), so =



=1(3 − 2)

(3)! ^3and thus

() = 1 + ^∞

=1



=1(3 − 2)

(3)! ^3. Both Maple and Mathematica are able to plot  if we define it this way, and Derive is able to produce a similar graph using a suitable partial sum of ().

Derive, Maple and Mathematica all have two initially known Airy functions, called AI·SERIES(z,m) and BI·SERIES(z,m) from BESSEL.MTH in Derive and AiryAi and AiryBi in Maple and Mathematica (just Ai and Bi in older versions of Maple). However, it is very difficult to solve for  in terms of the CAS’s Airy functions, although in fact () =

√3AiryAi() + AiryBi()

√3AiryAi(0) + AiryBi(0).

37. 2−1= 1 + 2 + ²+ 2³+ ⁴+ 2⁵+ · · · + ^2⁻²+ 2^2⁻¹

= 1(1 + 2) + ²(1 + 2) + ⁴(1 + 2) + · · · + ^2⁻²(1 + 2) = (1 + 2)(1 + ²+ ⁴+ · · · + ^2⁻²)

= (1 + 2)1 − ^2

1 − ² [by (11.2.3) with  = ²] → 1 + 2

1 − ² as  → ∞ by (11.2.4), when ||  1.

Also 2= 2−1+ ^2→ 1 + 2

1 − ² since ^2→ 0 for ||  1. Therefore, ^→ 1 + 2

1 − ² since 2and 2−1both approach 1 + 2

1 − ² as  → ∞. Thus, the interval of convergence is (−1 1) and () = 1 + 2

1 − ². 38. 4−1= 0+ 1 + 2²+ 3³+ 0⁴+ 1⁵+ 2⁶+ 3⁷+ · · · + ³^4⁻¹

=

0+ 1 + 2²+ 3³ 

1 + ⁴+ ⁸+ · · · + ^4⁻⁴

→ 0+ 1 + 2²+ 3³

1 − ⁴ as  → ∞

[by (11.2.4) with  = ⁴] for⁴  1 ⇔ ||  1. Also ^4, 4+1, 4+2have the same limits

(for example, 4= 4−1+ 0^4and ^4→ 0 for ||  1). So if at least one of ⁰, 1, 2, and 3is nonzero, then the interval of convergence is (−1 1) and () = 0+ 1 + 2²+ 3³

1 − ⁴ .

39. We use the Root Test on the series

^. We need lim

→∞



|^^| = || lim__→∞ ^

|^| =  ||  1 for convergence, or

||  1, so  = 1.

40. Suppose 6= 0. Applying the Ratio Test to the series

( − )^, we find that

 = lim

→∞



+1





 = lim→∞



+1( − )^+1

( − )^



 = lim→∞

| − |

|^+1|() = | − |

lim→∞|^+1| (if lim

→∞|^+1| 6= 0), so the series converges when | − |

lim→∞|^+1|  1 ⇔ | − |  lim_

→∞



 

+1



. Thus,  = lim→∞



 

+1



. If lim→∞



 

+1



 = 0

and | − | 6= 0, then () shows that  = ∞ and so the series diverges, and hence,  = 0. Thus, in all cases,

 = lim

→∞



 

+1



.

41. For 2    3,

^diverges and

^converges. By Exercise 11.2.85,

(+ ) ^diverges. Since both series converge for ||  2, the radius of convergence of

(+ ) ^is 2.

46. Suppose that the radius of convergence of the power series P cnxⁿ is R. What is the radius of convergence of the power seriesP cnx²ⁿ?

Solution: SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 169

42. Since

^converges whenever ||  ,

^2=



²

converges whenever²

   ⇔ || √

, so the second series has radius of convergence√.

11.9 Representations of Functions as Power Series

1. If () = ^∞

=0

^has radius of convergence 10, then ⁰() = ^∞

=1

^⁻¹also has radius of convergence 10 by Theorem 2.

2. If () = ^∞

=0

^converges on (−2 2), then

 ()  =  + ^∞

=0



 + 1^+1has the same radius of convergence (by Theorem 2), but may not have the same interval of convergence—it may happen that the integrated series converges at an endpoint (or both endpoints).

3. () = 1

 + 10 = 1 10

 1

1 − (−10)



= 1 10

∞

=0



− 10



or, equivalently, ^∞

=0(−1)^ 1

10^+1^. The series converges when 

10



  1, that is, when ||  10, so  = 10 and  = (−10 10).

4. () = 5

1 − 4² = 5

 1

1 − 4²



= 5^∞

=0

(4²)^= 5 ^∞

=0

4^^2. The series converges when4²  1 ⇔

||² ¹₄ ⇔ ||  ¹2, so  = ¹₂ and  =

−¹2¹₂.

5. () = 2 3 −  = 2

3

 1

1 − 3



= 2 3

∞

=0

  3



or, equivalently, 2^∞

=0

1

3^+1^. The series converges when 3



  1, that is, when ||  3, so  = 3 and  = (−3 3).

6. () = 4 2 + 3= 4

3

 1

1 + 23



= 4 3

 1

1 − (−23)



= 4 3

∞

=0



−2

3



=0(−1)^2^+2 3^+1^. The series converges when



−2

3



  1, that is, when ||  ³², so  = ³₂ and  =

−³2³₂.

7. () = ²

⁴+ 16 = ² 16

 1

1 + ⁴16



= ² 16

 1

1 − [−(2)]⁴



= ² 16

∞

=0



−  2

4

=0

(−1)^^4+2 2^4+4 . The series converges when



−  2

4  1 ⇒



 2



  1 ⇒ ||  2, so  = 2 and  = (−2 2).

8. () = 

2²+ 1 = 

 1

1 − (−2²)



=  ^∞

=0(−2²)^or, equivalently, ^∞

=0(−1)^2^^2+1. The series converges when

−2²

  1 ⇒ ²  ¹₂ ⇒ ||  1

√2, so  = 1

√2 and  =



− 1

√2 1

√2

 .

2