11.8 Power Series

(1)

Section 11.7 Strategy for Testing Series

45. Test the series for convergence or divergence.

∞

P

n=1 1 n^{1+ 1}n

. Solution:

SECTION 11.7 STRATEGY FOR TESTING SERIES ¤ 159

27.  ∞ 2

ln 

²  = lim

→∞



−ln 

 − 1





1

[using integration by parts] = 1. So^H ^∞

=1

ln 

² converges by the Integral Test, and since

 ln 

( + 1)³   ln 

³ = ln 

² , the given series ^∞

=1

 ln 

( + 1)³ converges by the Comparison Test.

28. Since

1





is a decreasing sequence, ^1≤ ^11= for all  ≥ 1, and ^∞

=1



² converges ( = 2  1), so ^∞

=1

^1

² converges by the Comparison Test. (Or use the Integral Test.)

29. ^∞

=1

= ^∞

=1(−1)^ 1

cosh  = ^∞

=1(−1)^. Now = 1

cosh   0, {^} is decreasing, and lim_

→∞= 0, so the series converges by the Alternating Series Test.

Or: Write 1

cosh = 2

^+ ^−  2

^ and ^∞

=1

1

^ is a convergent geometric series, so^∞

=1

1

cosh is convergent by the Comparison Test. So ^∞

=1(−1)^ 1

cosh is absolutely convergent and therefore convergent.

30. Let () =

√

 + 5. Then () is continuous and positive on [1 ∞), and since ⁰() = 5 −  2√

 ( + 5)²  0for   5, () is eventually decreasing, so we can use the Alternating Series Test. lim

→∞

√

 + 5 = lim

→∞

1

^12+ 5^−12 = 0, so the series

∞

=1(−1)^

√

 + 5 converges.

31. lim

→∞= lim

→∞

5^

3^+ 4^ = [divide by 4^] lim

→∞

(54)^

(34)^+ 1 = ∞ since lim_

→∞

3 4



= 0and lim

→∞

5 4



= ∞.

Thus, ^∞

=1

5^

3^+ 4^ diverges by the Test for Divergence.

32. lim

→∞



|^| = lim_

→∞



(!)^

^4



 = lim→∞

!

⁴ = lim

→∞



·  − 1

 · − 2

 ·  − 3

 · ( − 4)!



= lim

→∞



1 − 1





1 − 2





1 − 3





( − 4)!



= ∞,

so the series ^∞

=1

(!)^

^4 diverges by the Root Test.

33. lim

→∞



|^| = lim_

→∞

 

 + 1

²

= lim

→∞

1

[( + 1) ]^ = 1

lim→∞(1 + 1)^ = 1

  1, so the series ^∞

=1

 

 + 1

²

converges by the Root Test.

34. 0 ≤  cos² ≤ , so 1

 +  cos² ≥ 1

 +  = 1

2. Thus, ^∞

=1

1

 +  cos² diverges by comparison with ^∞

=1

1

2, which is a constant multiple of the (divergent) harmonic series.

35. = 1

^1+1 = 1

 · ^1, so let = 1

 and use the Limit Comparison Test. lim

→∞





= lim

→∞

1

^1 = 1  0 (see Exercise 6.8.63 [ET 4.4.63), so the series ^∞

=1

1

^1+1 diverges by comparison with the divergent harmonic series.

∞

P

n=1

(√ⁿ

2 − 1)ⁿ. Solution:

160 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 36. Note that (ln )^{ln }=

^{ln ln }ln 

=

^{ln }ln ln 

= ^{ln ln }and ln ln  → ∞ as  → ∞, so ln ln   2 for sufficiently large . For these  we have (ln )^{ln } ², so 1

(ln )^{ln }  1

². Since ^∞

=2

1

² converges [ = 2  1], so does

∞

=2

1

(ln )^{ln } by the Comparison Test.

37. lim

→∞



|^| = lim_

→∞(2^1− 1) = 1 − 1 = 0  1, so the series ^∞

=1

√

2 − 1^converges by the Root Test.

38. Use the Limit Comparison Test with = √^

2 − 1 and ^= 1. Then

lim→∞





= lim

→∞

2^1− 1 1 = lim

→∞

2^1− 1 1

= limH

→∞

2^1· ln 2 · (−1²)

−1² = lim

→∞(2^1· ln 2) = 1 · ln 2 = ln 2  0.

So since ^∞

=1

diverges (harmonic series), so does ^∞

=1

√

2 − 1.

Alternate solution: √^

2 − 1 = 1

2^(^−1)+ 2^(^−2)+ 2^(^−3)+ · · · + 2^1+ 1 [rationalize the numerator] ≥ 1 2, and since ^∞

=1

1 2 = 1

2

∞

=1

1

 diverges (harmonic series), so does ^∞

=1

√

2 − 1by the Comparison Test.

11.8 Power Series

1. A power series is a series of the form∞

=0^= 0+ 1 + 2²+ 3³+ · · · , where  is a variable and the ^’s are constants called the coefficients of the series.

More generally, a series of the form∞

=0( − )^= 0+ 1( − ) + ²( − )²+ · · · is called a power series in ( − ) or a power series centered at  or a power series about , where  is a constant.

2. (a) Given the power series∞

=0( − )^, the radius of convergence is:

(i) 0 if the series converges only when  =  (ii) ∞ if the series converges for all , or

(iii) a positive number  such that the series converges if | − |   and diverges if | − |  .

In most cases,  can be found by using the Ratio Test.

(b) The interval of convergence of a power series is the interval that consists of all values of  for which the series converges.

Corresponding to the cases in part (a), the interval of convergence is: (i) the single point {}, (ii) all real numbers; that is, the real number line (−∞ ∞), or (iii) an interval with endpoints  −  and  +  which can contain neither, either, or both of the endpoints. In this case, we must test the series for convergence at each endpoint to determine the interval of convergence.

3. If  = (−1)^^, then

lim→∞



+1





 = lim→∞



(−1)^+1( + 1)^+1 (−1)^^



 = lim→∞



(−1) + 1

 



 = lim→∞



1 + 1





||



= ||. By the Ratio Test, the

series ^∞

=1(−1)^^converges when ||  1, so the radius of convergence  = 1. Now we’ll check the endpoints, that is,

∞

P

n=1

(√ⁿ 2 − 1).

Solution:

160 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 36. Note that (ln )^{ln }=

^{ln ln }^{ln }

=

^{ln }^{ln ln }

= ^{ln ln }and ln ln  → ∞ as  → ∞, so ln ln   2 for sufficiently large . For these  we have (ln )^{ln } ², so 1

(ln )^{ln }  1

². Since ^∞

=2

1

² converges [ = 2  1], so does

∞

=2

1

(ln )^{ln } by the Comparison Test.

37. lim

→∞



|^| = lim__→∞(2^1− 1) = 1 − 1 = 0  1, so the series ^∞

=1

√

2 − 1^converges by the Root Test.

38. Use the Limit Comparison Test with = √^

2 − 1 and ^= 1. Then

lim→∞





= lim

→∞

2^1− 1 1 = lim

→∞

2^1− 1 1

= limH

→∞

2^1· ln 2 · (−1²)

−1² = lim

→∞(2^1· ln 2) = 1 · ln 2 = ln 2  0.

So since ^∞

=1

diverges (harmonic series), so does ^∞

=1

√

2 − 1.

Alternate solution: √^

2 − 1 = 1

2^(^−1)+ 2^(^−2)+ 2^(^−3)+ · · · + 2^1+ 1 [rationalize the numerator] ≥ 1 2, and since ^∞

=1

1 2 = 1

2

∞

=1

1

 diverges (harmonic series), so does ^∞

=1

√

2 − 1by the Comparison Test.

11.8 Power Series

1. A power series is a series of the form_∞

=0^= 0+ 1 + 2²+ 3³+ · · · , where  is a variable and the ^’s are constants called the coefficients of the series.

More generally, a series of the form_∞

=0( − )^= 0+ 1( − ) + ²( − )²+ · · · is called a power series in ( − ) or a power series centered at  or a power series about , where  is a constant.

2. (a) Given the power series_∞

=0( − )^, the radius of convergence is:

(i) 0 if the series converges only when  =  (ii) ∞ if the series converges for all , or

(iii) a positive number  such that the series converges if | − |   and diverges if | − |  .

In most cases,  can be found by using the Ratio Test.

(b) The interval of convergence of a power series is the interval that consists of all values of  for which the series converges.

Corresponding to the cases in part (a), the interval of convergence is: (i) the single point {}, (ii) all real numbers; that is, the real number line (−∞ ∞), or (iii) an interval with endpoints  −  and  +  which can contain neither, either, or both of the endpoints. In this case, we must test the series for convergence at each endpoint to determine the interval of convergence.

3. If  = (−1)^^, then

lim→∞



+1





 = lim→∞



(−1)^+1( + 1)^+1 (−1)^^



 = lim→∞



(−1) + 1

 



 = lim→∞



1 + 1





||



= ||. By the Ratio Test, the

series ^∞

=1(−1)^^converges when ||  1, so the radius of convergence  = 1. Now we’ll check the endpoints, that is,

1