Math Real Analysis

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HW1 Solution

Real Analysis II (Memorial University of Newfoundland)

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HW1 Solution

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 1

1. Determine the convergence/divergence of the following series. Find the sum where possible

(a) ∑

∞

n=1 1

(3n−2)(3n+1) ;

ANS: For all n ≥ 1. Let

an = 1

(3n − 2)(3n + 1) =

1

3 · (3n + 1) − (3n − 2) (3n − 2)(3n + 1)

= 1

3 · [

1

3n − 2 −

1

3n + 1

]

.

Therefore, the partial sum

sn = a1 + · · · + an =

1

3 · [

1 − 1

4

]

+ 1

3 · [

1

4 −

1

7

]

+ · · · + 1

3 · [

1

3n − 2 −

1

3n + 1

]

= 1

3 · [

1 − 1

3n + 1

]

.

So the limit of sn is 1/3 and hence

∞ ∑

n=1

1

(3n − 2)(3n + 1) =

1

3 .

(b) ∑

∞

n=1 1

n(n+2) ;

ANS: For all n ≥ 1. Let

an = 1

n(n + 2) =

1

2 · (n + 2) − n n(n + 2)

= 1

2 · [

1

n −

1

n + 2

]

.

Therefore, the partial sum

sn = a1 + · · · + an =

1

2

{

1 − 1

3 +

1

2 −

1

4 +

1

3 −

1

5 + · · · +

1

n − 1 −

1

n + 1 +

1

n −

1

n + 2

}

= 1

2 · [

1 + 1

2 −

1

n + 1 −

1

n + 2

]

.

So the limit of sn is 3/4 and hence

∞ ∑

n=1

1

n(n + 2) =

3

4 .

(c) ∑

∞

n=1 (−1)n−1

5n ;

ANS: This is a geometric series of ratio x = −1/5. Therefore, the sum ∞ ∑

n=1

(−1)n−1 5n

= 1/5

1 − (−1/5) =

1

6 .

(d) ∑

∞

n=1 (−1)n−1

n(n+2) ;

ANS: Let n ≥ 1 and then

bn = 1

n(n + 2) =

1

2 · [

1

n −

1

n + 2

]

.

Therefore, the term of ∑

∞

n=1 (−1)n−1

n(n+2) is an = (−1)n−1bn. Then

a2k = − 1

2 · [

1

2k −

1

2k + 2

]

, ∀k ≥ 1.

Similarly,

a2k−1 = 1

2 · [

1

2k − 1 −

1

2k + 1

]

, ∀k ≥ 1.

1

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Math 3001, Winter 2020 Homework 1

Therefore, the partial sum of ∑

∞

n=1 (−1)n−1

n(n+2) is

s2k = a1 + a2 + · · · + a2k = (a1 + a3 + · · · + a2k−1) + (a2 + a4 + · · · + a2k)

= 1

2

[(

1 − 1

2k + 1

)

− (

1

2 −

1

2k + 2

)]

= 1

2

[

1

2 −

1

2k + 1 +

1

2k + 2

]

.

Similarly, one has

s2k+1 = a1 + a2 + · · · + a2k + a2k+1 =

1

2

[

1

2 −

1

2k + 1 +

1

2k + 2

]

+ a2k+1

= 1

2

[

1

2 −

1

2k + 1 +

1

2k + 2

]

+ 1

2 · [

1

2k + 1 −

1

2k + 3

]

= 1

2

[

1

2 +

1

2k + 2 −

1

2k + 3

]

.

So one has lim k→∞

s2k = lim k→∞

s2k+1 = 1/4,

and limn→∞ sn = 1/4. That is,

∞ ∑

n=1

(−1)n−1 n(n + 2)

= 1

4 .

(e) ∑

∞

n=1[sin(n + 1) − sin(n)]; ANS: First of all, one gets the partial sum

sn = [sin(2)−sin(1)]+[sin(3)−sin(2)]+ · · ·+[sin(n+1)−sin(n)] = sin(n+1)−sin(1).

It then requires to prove that sin(n) does not have limit as follows.

First of all, let a = √ 3/2 and b = −

√ 3/2. To have sin(x) > a, one needs x ∈

(2kπ + π/3, 2kπ + 2π/3). As the distance of this interval is π/3 > 1, there is at least one integer locating in this interval. Let nk be any integer in x ∈ (2kπ+π/3, 2kπ+2π/3) (with n0 = 0), and one gets

sin(nk) > √ 3/2

for all k = 0, 1, 2, · · · Note that the sequence {sin(nk)} is bounded by 1 from above and by

√ 3/2 from below, then we can further choose a convergent subsequence sin(nkj )

such that, lim j→∞

sin(nkj ) = c, & c ≥ √ 3/2.

Similarly, to have sin(x) < b, one needs x ∈ (2kπ − 2π/3, 2kπ − π/3). As the distance of this interval is π/3 > 1, there is at least one integer locating in this interval. Let n′k be any integer in x ∈ (2kπ − 2π/3, 2kπ − π/3), and one gets

sin(n′k) < − √ 3/2

for all k = 0, 1, 2, · · · Note that the sequence {sin(n′k)} is bounded by −1 from below and by −

√ 3/2 from above, then we can further choose a convergent subsequence sin(n′kj )

such that, lim j→∞

sin(n′kj ) = d, & d ≤ − √ 3/2.

Assume that sin(n) has finite limit e, then any subsequence of sin(n) should have the same limit e; however, as above, we found two subsequence with absolutely different limits. Therefore, we can conclude that sin(n) does not have a finite limit (even the limit does not exist). Consequently, the desired series is divergent.

2

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Math 3001, Winter 2020 Homework 1

2. Find the value x so that

(a) ∑

∞

n=2 1

(3−x)n = 1/6;

ANS: This is a geometric series with ratio 1 3−x

and the first term 1 (3−x)2

. Therefore, if it has sum, then the sum is

s =

1 (3−x)2

1 − 1 3−x

= 1

(3 − x)2 − (3 − x) .

Let s = 1/6 which is equivalent (3 − x)2 − (3 − x) = 6. The solution for this equation is x = 0 and x = 5.

(b) ∑

∞

n=1 e −nx = 1.

ANS: This is a geometric series with ratio e−x and the first term e−x. Therefore, if it has sum, then the sum is

s = e−x

1 − e−x .

Let s = 1 which is equivalent ex = 2. The solution for this equation is x = ln 2.

3. Prove the following statements.

(a) If ∑

∞

n=1 an is convergent, then ∑

∞

n=1 2an is also convergent. Proof. As

∑

∞

n=1 an is convergent (say s), then the partial sum sn = a1 + · · · + an has limit s. Similarly, the partial sum of s1n = 2a1 + · · · + 2an = 2sn, and hence has limit 2s. Therefore,

∑

∞

n=1 2an is also convergent.

(b) If ∑

∞

n=1 an with an ≥ 0 is convergent, then ∑

∞

n=1 n+1 n an is also convergent.

Proof. As ∑

∞

n=1 an with an ≥ 0 is convergent, say to s, then the partial sum of ∑

∞

n=1 an has limit s. Similarly, the partial sum of

∑

∞

n=1 n+1 n an is

s1n = 2a1 + 3

2 a2 + · · ·

n + 1

n an ≤ 2sn

for all n ≥ 1. As an ≥ 0, hence s1n and sn are both monotone increasing. As sn ≤ s for all n, one has s1n ≤ 2s. The bounded monotone sequence theorem implies the convergence of

∑

∞

n=1 n+1 n an.

(c) If ∑

∞

n=1 an is convergent to s, then ∑

∞

n=1(an + an+2) is also convergent and calculate its sum. ANS: As the series

∑

∞

n=1 an is convergent to s, then the partial sn = a1 + · · · + an has limit s. Similarly, the partial sum of

∑

∞

n=1(an + an+2) is s 1 n = (a1 + a3) + (a2 + a4) +

· · · + (an + an+2) = 2sn − a1 − a2. Hence, the limit of s1n is 2s − a1 − a2, which implies the convergence of

∑

∞

n=1(an + an+2).

3

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