Section 11.6 Absolute Convergence and the Ratio and Root Tests

Section 11.6 Absolute Convergence and the Ratio and Root Tests

EX.21

Let an= n²+ 1 2n²+ 1

n

. lim

n→∞

p|an n| = lim

n→∞

n²+ 1 2n²+ 1 = 1

2 < 1, so the given series is absolutely convergent by the Root Test.

EX.27

Let a_n = (−1)ⁿ⁻¹1 · 3 · 5 · · · (2n − 1)

(2n − 1)! . Apply the Ratio Test,

n→∞lim    

a_n+1 a_n

= lim

n→∞

1 · 3 · · · (2n + 1) (2n + 1)!

(2n − 1)!

1 · 3 · · · (2n − 1) = lim

n→∞

1

2n = 0 < 1 Therefore the given series is absolutely convergent by the Ratio Test.

EX.28

First observe that the given series can be expressed as

∞

P

n=1

2 · 6 · · · (4n − 2)

5 · 8 · · · (3n + 2). Let a_n= 2 · 6 · · · (4n − 2) 5 · 8 · · · (3n + 2). Apply the Ratio Test,

n→∞lim    

a_n+1 a_n

= lim

n→∞

2 · 6 · · · (4n + 2) 5 · 8 · · · (3n + 5)

5 · 8 · · · (3n + 2)

2 · 6 · · · (4n − 2) = lim

n→∞

4n + 2 3n + 5 = 4

3 > 1 Therefore the given series is divergent by the Ratio Test.

EX.29

Since 2 · 4 · 6 · · · 2n

n! = 2ⁿ(1 · 2 · 3 · · · n)

n! = 2ⁿ, the given series is a divergent geometric series with

|r| = 2 > 1.

EX.31

Since a₁ = 2 > 0 and a_n > 0 ⇒ a_n+1 = 5n + 1

4n + 3a_n > 0 for all n > 1, we have a_n > 0 for all n ≥ 1 by mathematical induction and hence we can apply the Ratio Test to this series. lim

n→∞

a_n+1 an

=

n→∞lim

5n + 1 4n + 3 = 5

4 > 1, so the given series is divergent by the Ratio Test.

EX.38

(a) First we observe that for all m ≥ 1,

a_n+m= r_n+m−1a_n+m−1 = r_n+m−1r_n+m−2a_n+m−2 = · · · = r_n+m−1· · · r_n+1a_n+1 Since {r_n} is decreasing and r_n+1 < 1,

R_n = a_n+1+ a_n+2+ a_n+3+ a_n+4+ · · ·

= a_n+1+ r_n+1a_n+1+ r_n+2r_n+1a_n+1+ r_n+3r_n+2r_n+1a_n+1+ · · ·

= a_n+1(1 + r_n+1+ r_n+2r_n+1+ r_n+3r_n+2r_n+1+ · · · )

≤ an+1(1 + rn+1+ r²_n+1+ r³_n+1+ · · · )

= an+1

1 − r_n+1

(b) Since {r_n} is increasing, r_n ≤ L for all n ≥ 1. To prove this, suppose that there exist an integer N such that rN > L. Then

r_n> r_N > L for all n ≥ N + 1 ⇒ lim

n→∞r_n 6= L a contradiction. Thus, because L < 1

R_n = a_n+1(1 + r_n+1+ r_n+2r_n+1+ r_n+3r_n+2r_n+1+ · · · )

≤ a_n+1(1 + L + L²+ L³+ · · · )

= a_n+1 1 − L

EX.39

(a) s₅ = 1

1 · 2¹ + 1

2 · 2² + 1

3 · 2³ + 1

4 · 2⁴ + 1

5 · 2⁵ = 661

960 ≈ 0.6885 r_n = n2ⁿ

(n + 1)2ⁿ⁺¹ = n

2(n + 1) increases as n → ∞ and lim

n→∞r_n = 1

2. By the result of Problem 38 (b), the error in using s₅ is

R₅ ≤ a₆

1 − 1/2 = 1/(6 · 2⁶)

1/2 = 1

192 ≈ 5.208 × 10⁻³ (b) 1/(n + 1)2ⁿ⁺¹

1 − 1/2 < 0.00005 ⇒ 20000 ≤ (n + 1)2ⁿ From the following table, we know that the above inequality holds whenever n ≥ 11.

n 8 9 10 11 12

(n + 1)2ⁿ 2304 5120 11264 24576 53244 Choose n = 11. The sum of the series is approximated by s₁₁=

11

P

n=1

1

n2ⁿ ≈ 0.693109 2

EX.43

(a) P a_n is absolutely convergent ⇒ P a_n is convergent. Since P a⁺_n/P a⁻_n is the sum/difference of two convergent sequences times a constant (1/2), it is convergent as well.

(b) Notice that |a_n| = 2a⁺_n− a_n = a_n− 2a⁻_n. IfP a⁺_n orP a⁻_n converges, thenP |a_n|, as the difference of two convergent sequences, converges as well, which contradicts to the fact that P an is not absolutely convergent. Therefore both P a⁺_n and P a⁻_n are divergent.

EX.44

Following the hint, for any real number r we will construct a rearranged sequence {ˆan} of {an} such that it sums up to r. Here we will assume that r ≥ 0. For r < 0, just interchange the role of a⁺_n and a⁻_n in this proof.

First, let ˆsn denote the partial sum of the sequence {ˆan} and define ˆs0 = 0.

Next, take the following steps recursively to construct {ˆan}:

(1) Let σ₀ = l⁺₀ = l⁻₀ = 0.

(2) For m ≥ 1

Case I. If m is odd, m = 2k − 1, find l⁺_k such that

ˆ s_σm−1+

l⁺_k−1

X

n=l⁺_k−1+1

a⁺_n ≤ r < ˆs_σm−1+

l⁺_k

X

n=l⁺_k−1+1

a⁺_n (*)

Let σ_m = σ_m−1 + (l_k⁺− l_k−1⁺ ).

For i = 1, 2, · · · , l⁺_k − l⁺_k−1, assign ˆaσm−1+i = a_l⁺

k−1+i. Case II. If m is even, m = 2k, find l_k⁻ such that

ˆ s_σm−1+

l⁻_k−1

X

n=l⁻_k−1+1

a⁻_n ≥ r > ˆs_σm−1+

l⁻_k

X

n=l⁻_k−1+1

a⁻_n (**)

Let σ_m = σ_m−1 + (l_k⁻− l_k−1⁻ ).

For i = 1, 2, · · · , l⁻_k − l⁻_k−1, assign ˆa_σm−1+i = a_l⁻

k−1+i.

where the zeros terms in {a⁺_n} and {a⁻_n} are discarded and the subscripts are relabeled.

Note that the existence of l_k⁺ and l_k⁻ in (*) and (**) is guaranteed by the fact that both P a⁺_n and P a⁻_n are divergent which is proved in Problem 43 (b). The other thing to notice is that the constructed sequence {ˆa_n} is indeed a rearrangement of {ˆa_n}, that is, this provides an 1-to-1 and onto map from {a_n} to {ˆa_n} (check it yourself!).

Finally, let us check whether

∞

P

n=1

ˆ

a_n = r. Given σ_m−1 < n ≤ σ_m, by the construction process we have the error

|ˆs_n− r| ≤ max(|ˆs_σm−1 − r|, |ˆs_σm − r|) ≤ max(|ˆa_σm−1|, |ˆa_σm|) Since P a_n is conditionally convergent, lim

n→∞a_n = 0. So lim

n→∞ˆa_n = 0 and hence lim

n→∞|ˆs_n − r| = 0.

Therefore

∞

P

n=1

ˆ

a_n = r and we complete the proof.

The following figure should help you understand the construction process:

4

Figure 1: *

EX.45

(a) Suppose that P n²an converges. Then lim

n→∞n²an = 0 ⇒ lim

n→∞|n²an| = lim

n→∞n²|an| = 0. By applying the Comparison Test for the seriesP |an| with the convergent p-seriesP 1

n² (p = 2 > 1), we have lim

n→∞

|an|

1/n² = lim

n→∞n²|a_n| = 0. This implies that P |a_n| is convergent and hence P a_n is absolutely convergent, a contradiction. Thus, P n²a_n is divergent.

(b) Let a_n = (−1)ⁿ⁻¹1

n. By the Alternating Series Test, P a_n is conditionally convergent. But P na_n=P(−1)ⁿ⁻¹ is divergent because the limit of its partial sum does not exist.