Math Real Analysis

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HW4-Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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HW4-Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 4 Due: Feb 24, 2020

(a): Determine the convergence/divergence of the following series.

1. ∑∞

n=1 1

nln n ;

ANS: Notice that, when n ≥ e2, say n ≥ 8, then ln n ≥ 2. Therefore, nln n ≥ n2 for all n ≥ 8. Equivalently,

1

nln n ≤

1

n2 , ∀n ≥ 8.

We know that ∑

1/n2 is a convergent series (p-series with p = 2). By comparison test (inequality form), one has,

∑∞ n=1

1 nln n

is convergent.

2. ∑∞

n=1 n!3n

nn ;

ANS: Notice that we have n!, and we should try ratio test first. That is,

lim n→∞

(n + 1)!3n+1

(n + 1)n+1 ×

nn

n!3n = lim

n→∞ 3(1 + 1/n)−n = 3/e > 1.

Therefore, ∑∞

n=1 n!3n

nn is divergent.

3. ∑∞

n=1 n

3 √

n ;

ANS: Note that

lim n→∞

n3

3 √ n = lim

x→∞ x6

3x = 0

by L’-Hopital. Basically, we compare ∑∞

n=1 n

3 √

n with

∑∞ n=1 1/n

2, one gets the convergence of ∑∞

n=1 n

3 √

n .

4. ∑∞

n=1 n(n+1)

4n ;

ANS: We can use both ratio test and root test. For example, we use ratio test as follows:

lim n→∞

(n + 2)(n + 1)

4n+1 ×

4n

n(n + 1) = 1/4 < 1

and then ∑∞

n=1 n(n+1)

4n is convergent.

5. ∑∞

n=1 (2n)!

22n(n!)2 ;

ANS: It involves with n!, and we try ratio test as follows.

lim n→∞

(2n + 2)!

22n+2((n + 1)!)2 ×

22n(n!)2

(2n)! = lim

n→∞ 2n + 1

2n + 2 = 1.

In this case, we need to use Raabe’s test as follows:

r = lim n→∞

n

[

1 − 2n + 1

2n + 2

]

= lim n→∞

n

2n + 2 = 1/2 < 1

and hence ∑∞

n=1 (2n)!

22n(n!)2 is divergent.

6. ∑∞

n=1 3·5·7···(2n+1) 1·4·7···(3n−2);

ANS: We should try the ratio test first as follows:

L = lim n→∞

3 · 5 · 7 · · · (2n + 3) 1 · 4 · 7 · · · (3n + 1)

× 1 · 4 · 7 · · · (3n − 2) 3 · 5 · 7 · · · (2n + 1)

= lim n→∞

2n + 3

3n + 1 = 2/3 < 1,

therefore, the series is convergent.

1

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Math 3001, Winter 2020 Homework 4 Due: Feb 24, 2020

7. ∑∞

n=1 an

1+a2n ;

ANS: When |a| < 1, one knows that ∣

∣

∣

∣

an

1 + a2n

∣

∣

∣

∣

< |a|n

and ∑

|a|n is a converge geometric series with ratio |a| < 1. Hence comparison test (inequality form) implies the convergence of

∑∞ n=1

an

1+a2n for the case of |a| < 1.

If a = 1, one gets ∑∞

n=1 an

1+a2n is divergent since a

n

1+a2n = 1/2 for all n.

When a = −1, an 1+a2n

= (−1)n/2, which is divergent since the general term does not have limit 0. For |a| > 1, one has

∣

∣

∣

∣

an

1 + a2n

∣

∣

∣

∣

< |a|n |a|2n

=

(

1

|a|

)n

for all n. Note that 1/|a| < 1 and by comparison test (inequality form), one gets the convergence of ∑

(

1 |a|

)n implies the convergence of

∑∞ n=1

an

1+a2n .

8. ∑∞

n=1 nn−1

(2n2+n+1) n−1 2

;

ANS: Let us use the root test as follows:

(

nn−1

(2n2 + n + 1) n−1 2

) 1

n−1

= n

(2n2 + n + 1) 1

2

= 1√ 2 < 1

and we have convergence.

9. ∑∞

n=1 2 −n−(−1)n;

ANS: Notice that by −n − (−1)n ≤ 1 − n, and hence 2−n−(−1)n ≤ 21−n. Also note that the geometric series

∑

21−n is convergent since the ratio is 1/2, hence the comparison test (inequality form) implies the convergence of

∑∞ n=1 2

−n−(−1)n.

10. ∑∞

n=1

[

(2n−1)!! (2n)!!

]p ; here (2n)!! = 2 · 4 · 6 · · · (2n) and (2n − 1)!! = 1 · 3 · 5 · · · (2n − 1);

ANS: We should use the ratio test first as follows:

lim n→∞

[

(2n + 1)!!

(2n + 2)!!

]p

× [

(2n)!!

(2n − 1)!!

]p

= lim n→∞

(

2n + 1

2n + 2

)p

= 1

and hence the ratio test does not work here.

We need to use the Raabe’s test as follows:

r = lim n→∞

n

[

1 − (

2n + 1

2n + 2

)p]

= p/2. (1)

Hence, if p > 2, the series is convergent as r > 1. If p < 2, the series is divergent as r < 1. For r = 1, we need to use the Gauss Test.

2

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Math 3001, Winter 2020 Homework 4 Due: Feb 24, 2020

Now we show the calculation of equation (1) as follows.

r = lim n→∞

n

[

1 − (

2n + 1

2n + 2

)p]

= lim n→∞

[

1 − (

2+1/n 2/n+2

)p]

1/n

= lim x→0

[

1 − (

2+x 2x+2

)p]

x

= lim x→0

(2x + 2)p − (2 + x)p x(2x + 2)p

= lim x→0

2p(2x + 2)p−1 − p(2 + x)p−1 (2x + 2)p + 2px(2x + 2)p−1

= p/2

(Hint: Always do algebra to simplify your formula, in particular to avoid the quotient rules).

For p = 2, one has

n

[

1 − (

2n + 1

2n + 2

)2 ]

= n

[

(2n + 2)2 − (2n + 1)2 (2n + 2)2

]

= 4n2 + 3n

4n2 + 8n + 4

= 1 − 1

n ×

5n2 + 4n

4n2 + 8n + 4

which satisfies the conditions for Gauss test with c = 1 and the bounded sequence βn = 5n2+4n

4n2+8n+4 (βn has limit 5/4 and hence it is bounded). By Gauss test, the series is divergent.

11. ∑∞

n=1 α·(α+1)···(α+n−1)

n!nβ , (α > 0, β > 0);

ANS: We shall use the ratio test first as follows:

lim n→∞

α · (α + 1) · · · (α + n − 1)(α + n) (n + 1)!(n + 1)β

× n!nβ

α · (α + 1) · · · (α + n − 1) = lim

n→∞ (α + n)nβ

(n + 1)β+1 = 1

for all α, β > 0. Hence the ratio test does not work and we need the Raabe’s test as follows:

r = lim n→∞

n

[

1 − (α + n)nβ

(n + 1)β+1

]

= lim n→∞

[

1 − (α/n+1) (1/n+1)β+1

]

1/n

= lim x→0

[

1 − (αx+1) (x+1)β+1

]

x

= lim x→0

(x + 1)β+1 − (αx + 1) x(x + 1)β+1

= lim x→0

(β + 1)(x + 1)β − α (x + 1)β+1 + x(β + 1)(x + 1)β

= β + 1 − α.

3

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Math 3001, Winter 2020 Homework 4 Due: Feb 24, 2020

Hence the series is converge if r = β + 1 − α > 1, i.e., β > α; it is divergent if r = β + 1 − α < 1, i.e., β < α.

In the case of β = α, the Raabe’s test does not work. We will use the Gauss test as follows. Now let us calculate the following limit (and most of the calculation follows from those in Raabe’s test):

r′ = lim n→∞

n

{

1 − n [

1 − (α + n)nα

(n + 1)α+1

]}

= lim x→0

x(x + 1)α+1 − (x + 1)α+1 + 1 + αx x2(x + 1)α+1

= (α − 2)(α + 1)

2 ,

by using L-Hoptal twice. Hence the sequence

βn = n

{

1 − n [

1 − (α + n)nα

(n + 1)α+1

]}

is of course a bounded sequence. Equivalently, one has

n

[

1 − (α + n)nα

(n + 1)α+1

]

= 1 − tn n .

By Gauss test, the series is divergent.

12. ∑∞

n=1(1 − 1/n)n 2

;

ANS: By the root test, we have

lim n→∞

[(1 − 1/n)n2]1/n = lim n→∞

(1 − 1/n)n = e−1 < 1

and hence the series is convergent.

(b): Assume that the positive series ∑∞

n=1 an and ∑∞

n=1 bn are both divergent. Discuss the conver- gence/divergence of the following series:

1. ∑∞

n=1 min{an, bn};

ANS: This can be divergent or convergent depends on an and bn. For instance, if an = bn then ∑∞

n=1 min{an, bn} is just ∑

an and diverges.

Now if we let a2k = 1/k 2 and a2k−1 = k, for k = 1, 2, · · · , one can check that lim an is not 0 and

hence ∑

an diverges. Let b2k−1 = 1/k 2 and b2k = k, for k = 1, 2, · · · , one can check that lim bn is

not 0 and hence ∑

bn diverges.

However, min{an, bn} = 1/k2 for n = 2k and n = 2k − 1. Hence ∑∞

n=1 min{an, bn} = 2 ∑

1/k2 and convergent.

2. ∑∞

n=1 max{an, bn};

ANS: Note that 0 ≤ an ≤ max{an, bn} and hence the divergence of ∑

an implies the divergence of ∑∞

n=1 max{an, bn}.

4

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Math 3001, Winter 2020 Homework 4 Due: Feb 24, 2020

(c): Prove: if ∑

an converges and ∑

a2n diverges, then ∑

an is converges conditionally. (Hint: prove by contradiction).

Proof. Since ∑

an converges then limn→∞ an = 0. This further implies that, |an| < M for all n and for some M.

Method1: Since |an|2 = |an||an| < M|an|, by the comparison test (inequality form), ∑

|an|2 diverges implies that

∑

|an| also diverges. Hence ∑

an converge conditionally.

Method 2: Assume that ∑

|an| converges, then again by |an|2 = |an||an| < M|an| and the comparison test (inequality form),

∑

a2n converges. This is a contradiction with our assumption. Hence ∑

|an| diverges and

∑

an converge conditionally.

5

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