Section 11.2 Series

Section 11.2 Series

EX.15

(a) lim

n→∞a_n= lim

n→∞

2n

3n+1 = ²₃, so {a_n} is convergent.

(b) ∵ lim

n→∞a_n= ²₃ 6= 0, so

∞

P

n=1

a_n is divergent.

EX.33

∵ lim

n→∞a_n= lim

n→∞

√n

2 = lim

n→∞2ⁿ¹ = 2⁰ = 1 6= 0, so

∞

P

n=1

√n

2 is divergent.

EX.41

∞

P

n=1 1

eⁿ is a geometric series with ratio r = ¹_e, and |r| = ¹_e < 1.

So,

∞

P

n=1 1 eⁿ =

1 n

1−_n¹ = _e−1¹ . And by example 7, we know that

∞

P

n=1 1

n(n+1) = 1. Thus,

∞

P

n=1

(_e¹n +_n(n+1)¹ ) =

∞

P

n=1 1 eⁿ +

∞

P

n=1 1

n(n+1) = _e−1¹ + 1 = _e−1^e

EX.48

Let s_n =

n

P

i=2 1

i(i−1)(i+1) =

n

P

i=2

(−¹_i + _i−1^1/2 + ^1/2_i+1) = ¹₂

n

P

i=2

(_i−1¹ − ²_i +_i+1¹ )

= ¹₂[(¹₁ −²₂ +¹₃) + (¹₂ − ²₃ + ¹₄) + (¹₃ − ²₄ +¹₅) + (¹₄ −²₅ + ¹₆) + · · · +(_n−3¹ − _n−2² +_n−1¹ ) + (_n−2¹ − _n−1² + ¹_n) + (_n−1¹ − ²_n+ _n+1¹ )]

= ¹₂(¹₁ − ²₂ +¹₂ + ¹_n− _n² + _n+1¹ ) = ¹₄ − _2n¹ + _2n+2¹ Thus,

∞

P

n=2 1

n³−n = lim

n→∞s_n= lim

n→∞(¹₄ − _2n¹ + _2n+2¹ ) = ¹₄

EX.49

(a) Many people may guess that x < 1. But in fact, x = 1.

(b) x = 0.99999... = ₁₀⁹ + ₁₀₀⁹ + ₁₀₀₀⁹ + ₁₀₀₀₀⁹ + · · · =

∞

P

n=1 9

10ⁿ, which is a geometric series with a₁ = 0.9 and r = ₁₀¹. Its sum is _1−0.1^0.9 = 1, so x = 1.

1

EX.70

(a) The residual concentration just before the (n + 1)st injection is De^−aT + De^−a2T + De^−a3T + · · · + De^−anT = ^De−aT_1−e^(1−e−aT^−anT)

(b) The limiting pre-injection concentration is

n→∞lim

De^−aT(1−e^−anT)

1−e^−aT = ^De_1−e^−aT−aT⁽¹⁻⁰⁾ = _eaT^D−1

(c) _eaT^D−1 ≥ C ⇒ D ≥ C(e^aT − 1), so the minimal dosage is D = C(e^aT − 1).

EX.71

(a) Sn = D + Dc + Dc²+ · · · + Dcⁿ⁻¹= ^D(1−c_1−cⁿ⁾ (b) lim

n→∞S_n = lim

n→∞

D(1−cⁿ)

1−c = _1−c^D lim

n→∞(1 − cⁿ) = _1−c^D (∵ 0 < c < 1 ⇒ lim

n→∞cⁿ= 0)

= ^D_s = kD. If c = 0.8, then s = 0.2 ⇒ k = ¹_s = 5

EX.76

The area between y = xⁿ⁻¹ and y = xⁿ for 0 ≤ x ≤ 1 is R1

0(xⁿ⁻¹− xⁿ)dx = [^x_nⁿ − ^x_n+1ⁿ⁺¹]¹₀ = ¹_n− _n+1¹ = _n(n+1)¹

we can see as n → ∞, the sum of the areas between the successive curves approches the area of the unit square.

So,

∞

P

n=1 1

n(n+1) = 1.

EX.80

Suppose

∞

P

n=1

a_n is convergent, then lim

n→∞a_n = 0 ⇒ lim

n→∞

1 an 6= 0,

∴

∞

P

n=1 1

an is divergent.

EX.83

Suppose that P(a_n+ b_n) and P a_n are convergent. Then by Theorem 8(iii), P[(an+ b_n) − a_n] =P bn is also convergent. But this is a contradiction.

∴P(a_n+ b_n) is divergent.

2

EX.84

No. For example, let a_n= n & b_n= −n. Then bothP a_nandP b_nare divergent, butP(a_n+b_n) =P 0 is convergent to 0.

EX.88

(a) If a₁ = 1, a₂ = 2, then the limit seem to be ⁵₃. If a1 = 2, a2 = 3, then the limit seem to be ⁸₃. If a₁ = 4, a₂ = 1, then the limit seem to be 2.

If a1 = 1, a2 = 4, then the limit seem to be 3.

So, we would guess that the limit is ^a1+2a₃ ².

(b) an+1− an = ¹₂(an+ an−1) − an = −¹₂(an− an−1) = · · · = (−¹₂)ⁿ⁻¹(a2− a1).

And a_n = a₁+ (a₂− a₁) + (a₃− a₂) + · · · + (a_n−1− a_n−2) + (a_n− a_n−1)

= a1+

n−1

P

k=1

(ak+1− ak) = a1+

n−1

P

k=1

(−¹₂)^k−1(a2 − a1).

So, lim

n→∞an= a1+ (a2− a1)

∞

P

k=1

(−¹₂)^k−1 = a1+ (a2− a1)[_1−(−1/2)¹ ] = ^a1+2a₃ ²

EX.89

(a) s₁ = _1·2¹ = ¹₂, s₂ = ¹₂ +_1·2·3¹ = ⁵₆, s₃ = ⁵₆ + _1·2·3·4¹ = ²³₂₄, s₄ = ²³₂₄+ _{1·2·3·4·5}¹ = ¹¹⁹₁₂₀ The denominator are (n + 1)!, so we guess s_n= ^(n+1)!−1_(n+1)! .

(b) For n = 1, s₁ = ¹₂ = ^2!−1_2! , it holds for n = 1. Assume s_k= ^(k+1)!−1_(k+1)! . Then

sk+1 = ^(k+1)!−1_(k+1)! +_(k+2)!^k+1 = (k+2)!−(k+2)+k+1

(k+2)! = ^(k+2)!−1_(k+2)!

Thus, it holds for n = k + 1. So our guess is correct by induction.

(c)

∞

P

n=1 n

(n+1)! = lim

n→∞sn = lim

n→∞

(n+1)!−1

(n+1)! = lim

n→∞[1 − _(n+1)!¹ ] = 1

3