Math Real Analysis

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HW9Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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HW9Solution - Assignment solutions

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 9

You should do this homework assignment. There will be some problems in the final exam similar to those in this homework assignment! (a): Prove that

∫ π/2

0

∞ ∑

n=1

cos(nx)

n2 =

∞ ∑

k=1

(−1)k−1 (2k − 1)3

.

ANS: First of all, by M-test, one can prove that ∑∞

n=1 cos(nx)

n2 is uniformly convergent

as follows

| cos(nx)

n2 | ≤ 1/n2,

∑

1/n2 converges.

For each n, the function cos(nx)

n2 is continuous on [0, π/2]. Hence it is Riemann inte-

grable. By integration term-by-term, one gets

∫ π/2

0

∞ ∑

n=1

cos(nx)

n2 =

∞ ∑

n=1

∫ π/2

0

cos(nx)

n2 =

∞ ∑

n=1

sin(nx)

n3 |x=π/2x=0 .

Notice that, if n is even, then sin(nπ/2) = 0. If n = 4k + 1 = 2(2k) + 1, one has sin(nπ/2) = 1, and if n = 4k + 3 = 2(2k + 1) + 1, one gets sin(nπ/2) = −1. Putting this back, one gets the desired results. (b): Prove that

d

dx

∞ ∑

n=1

cos(nx)

n2 = −

∞ ∑

n=1

sin(nx)

n , ∀x ∈ (0, 2π).

First of all, ∞ ∑

n=1

cos(nx)

n2

is uniformly convergent on x ∈ (0, 2π). Second, each function cos(nx) n2

is differentiable

with derivative −sin(nx) n

. Moreover, by Abel’s or Dirichlet test, (see my notes for this example),

− ∞ ∑

n=1

sin(nx)

n

is uniformly convergent on any closed interval [a, b] ⊂ (0, 2π). Hence, by differentiable term-by-term, we gets the results holds on any closed interval [a, b] ⊂ (0, 2π). How- ever, for any x ∈ (0, 2π), one can always find δ > 0, such that [x−δ, x+δ] ⊂ (0, 2π). Hence,

d

dx

∞ ∑

n=1

cos(nx)

n2 = −

∞ ∑

n=1

sin(nx)

n , ∀x ∈ (0, 2π).

(b): Let f(x) = ∑∞

n=1 e−nx

1+n2 .

1. Prove that f(x) is continuous on I = [0, ∞).

First of all, notices that

e−nx

1 + n2 ≤ 1/n2, ∀x ≥ 0

and hence by M-test, it is uniformly convergent. Therefore, f(x) is continuous since each function e

−nx

1+n2 is continuous.

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Math 3001, Winter 2020 Homework 9

2. Prove that

f ′(x) = ∞ ∑

n=1

−ne−nx 1 + n2

, ∀x > 0.

First of all, each function e −nx

1+n2 is differentiable with derivative −ne

−nx

1+n2 .

Second, we let an(x) = e −nx, one can verify that

∑

an(x) is uniformly conver- gent for all x ∈ [a, ∞) with a > 0. In fact, it is a geometric series with ratio e−x ≤ e−a < 1 for all x ∈ [a, ∞). Let bn(x) = n1+n2 . It is decreasing (to 0). Hence, by Abel’s or Dirichlet test,

∞ ∑

n=1

−ne−nx 1 + n2

is uniformly convergent on any interval [a, ∞). Hence, by differentiable term- by-term, we gets the results holds on any closed interval [a, ∞). However, for any x ∈ (0, ∞), one can always find a > 0, such that x ∈ [a, ∞). Hence,

f ′(x) = ∞ ∑

n=1

−ne−nx 1 + n2

, ∀x > 0.

(d): Find the radius of convergence of the following series.

1. ∑∞

n=1 (2n)!!xn

(2n+1)!! .

ANS: Let bn = (2n)!!xn

(2n+1)!! . To have

∑

bn convergent, one needs

lim | bn+1 bn

| = lim | (2n + 2)!!xnx

(2n + 3)!! ×

(2n + 1)!!

(2n)!!xn | = lim |x|

2n + 2

2n + 3 = |x| < 1.

Hence the radius of convergence is R = 1.

2. ∑∞

n=1(1 + 1/n) −n2xn.

ANS: Let bn = (1 + 1/n) −n2xn. To have

∑

bn convergent, one needs

lim |bn|1/n = lim |(1 + 1/n)−nx| = lim |x|/e < 1.

Equivalently, ∑

bn converges if |x| < e. Hence the radius of convergence is R = e.

3. ∑∞

n=1(1 + 1/n) n2x2n.

ANS: Let bn = (1 + 1/n) n2x2n. To have

∑

bn convergent, one needs

lim |bn|1/n = lim |(1 + 1/n)nx2| = lim ex2 < 1.

Equivalently, ∑

bn converges if |x| < √

1/e. Hence, the radius of convergence

is R = √

1/e.

Method 2: You should notice that a2n−1 = 0 and a2n = (1 + 1/n) n2. Hence,

you should use

1/R = lim sup |an|1/n = lim[(1 + 1/n)n 2

]1/(2n) = e1/2.

Hence, R = e−1/2.

Method 3: Let y = x2. Then the series is ∑

(1 + 1/n)n 2

yn. For this series, an = (1 + 1/n)

n2, and its radius of convergence is R′ = 1/e. Notice y = x2. This implies that the original series is convergent if x2 < 1/e, and this gives R = e−1/2.

2

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Math 3001, Winter 2020 Homework 9

4. ∑∞

n=1 xn

2

2n .

ANS: Let bn = xn

2

2n . To have

∑

bn convergent, one needs

lim |bn|1/n = lim |x|n 2

< 1.

Equivalently, ∑

bn converges if |x| ≤ 1 (otherwise, |x|n → ∞ if |x| > 1. Hence, the radius of convergence is R = 1.

Method 2: You should notice that an2 = 2 −n and am = 0 for m 6= n2. Hence,

you should use 1/R = lim sup(2n)1/n

2

= lim 21/n = 1.

Hence, R = 1.

Method 3: Let y = xn. Then the series is ∑

yn/2n. For this series, an = 2 −n,

and its radius of convergence is R′ = 2. Notice y = xn. This implies that the original series is convergent if xn < 2, and this gives R = 1.

(c): Find the interval of convergence of the following series.

1. ∑∞

n=1 xn

3 √

n .

ANS: Let bn = xn

3 √

n . To have ∑

bn convergent, one needs

lim | xn

3 √ n |1/n = lim

|x| 31/

√ n = |x| < 1.

Hence the radius of convergence is R = 1.

At x = 1, the series is ∑

1 3 √

n , which is convergent. (See the midterm exam for similar questions). You can use integral test, Cauchy condensation test. I will complete this as follows. Note that ln(3

√ n) =

√ n ln 3 >

√ n ≥ 2 ln n if

(say) n > e8). Hence that for all n > e8, 3 √ n > n2 and hence 1/3

√ n ≤ 1/n2.

The later one forms a convergent series, and by comparison test, ∑

1 3 √

n is

convergent. This also implies that ∑ (−1)n

3 √

n is convergent (in fact, the case of x = 1 implies it is absolutely convergent). Hence the interval of convergence is [−1, 1].

2. ∑∞

n=1 xn√ n2+1

.

ANS: Let bn = xn√ n2+1

. To have ∑

bn convergent, one needs

lim | xn+1

√

(n + 1)2 + 1 ×

√ n2 + 1

xn | = |x| < 1.

Hence the radius of convergence is R = 1.

At x = 1, the series is ∑

1√ n2+1

, which is comparable with ∑

1/n, and hence it is divergent.

At x = −1, the series is ∑ (−1)n

√ n2+1

. It is alternating series with an = 1√

n2+1 ,

which is decreasing to 0. Hence Lebnitz test implies that it is convergent.

In conclusion, the interval of convergence is [−1, 1).

3. ∑∞

n=1 (2x)n

n!

ANS: Let bn = (2x)n

n! . To have

∑

bn convergent, one needs

lim | (2x)n+1

(n + 1)! ×

n!

(2x)n | = lim

|2x| n + 1

< 1.

3

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Math 3001, Winter 2020 Homework 9

Hence the radius of convergence is R = ∞ as for any fixed x ∈ (−∞, ∞),

lim | (2x)n+1

(n + 1)! ×

n!

(2x)n | = lim

|2x| n + 1

= 0 < 1.

In conclusion, the interval of convergence is (−∞, ∞).

4. ∑∞

n=1 (ln n)xn

2n .

ANS: Let bn = (ln n)xn

2n . To have

∑

bn convergent, one needs

lim | (ln n)xn

2n |1/n = lim

|x| 2 (ln n)1/n = |x|/2 < 1.

Hence the radius of convergence is R = 2. Here, to calculate lim(ln n)1/n, one needs

lim(ln n)1/n = exp(lim ln ln n

n ) = exp( lim

y→∞

ln ln y

y ) = exp( lim

y→∞

1

y ln y ) = exp(0) = 1.

At x = 2, the series is ∑

ln n, which is divergent, as the general term, ln n, not going to 0. Similarly, one gets the divergence at x = −2. In conclusion, the interval of convergence is (−2, 2).

5. ∑∞

n=1 n √ nxn.

ANS: Let bn = n √ nxn. To have

∑

bn convergent, one needs

lim |n √ nxn|1/n = |x| lim n1/

√ n = |x| < 1.

Hence the radius of convergence is R = 1. (Here, to calculate lim n1/ √ n, one

needs

lim n1/ √ n = exp(lim

ln n√ n ) = exp( lim

y→∞

2 ln y

y ) = exp( lim

y→∞

2

y ) = exp(0) = 1.

by letting n = y2).

At x = 1, the series is ∑

n √ n, which is divergent, as the general term, n

√ n,

not going to 0. Similarly, one gets the divergence at x = −1. In conclusion, the interval of convergence is (−1, 1).

(d): Find the following sum. (At least one big problem in the final exam is similar to these problems).

1. s(x) = ∑∞

n=1 n(n + 1)x n on x ∈ (−1, 1).

ANS: First of all, one can verify that the radius of convergence is R = 1, by

R = lim n(n + 1)

(n + 2)(n + 1) = 1.

At x = 1, the series is ∑

n(n + 1), which is divergent as the general term, n(n + 1), is not going to 0. Similarly, you get divergence at x = −1. So the interval of convergence is (−1, 1). In the interval (−1, 1), one can have

s1(x) =

∫ x

0

s(t) dt = ∞ ∑

n=1

nxn+1 = x2 ∞ ∑

n=1

nxn−1.

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Math 3001, Winter 2020 Homework 9

Let

s2(x) = s1(x)/x 2 =

∞ ∑

n=1

nxn−1.

In (−1, 1), one can calculate

s3(x) =

∫ x

0

s2(t) dt = ∞ ∑

n=1

xn = x

1 − x .

By the fundamental theorem of Calculus, one gets

s2(x) = (s3(x)) ′ =

1

(1 − x)2 ⇒ s1(x) = x2s2(x) =

x2

(1 − x)2 .

By the Fundamental theorem of Calculus again, one gets

s(x) = (s1(x)) ′ =

2x

(1 − x)3 .

2. s = ∑∞

n=1 1

n3n .

ANS: First of all, one consider the following series

s(x) = ∞ ∑

n=1

xn

n .

One can verify that the radius of convergence is R = 1, by

R = lim n + 1

n = 1.

At x = 1, the series is ∑

1/n, which is divergent. At x = −1, it is alternat- ing harmonic and hence convergent. Therefore, the convergence of interval is [−1, 1). Hence the series s =

∑∞ n=1

1 n3n

= s(1/3) is convergent.

In the interval (−1, 1), one can have

s1(x) = s ′(x) =

∞ ∑

n=1

xn−1 = 1

1 − x .

Hence s(x) = ∫ x

0 s1(t) dt = − ln(1−x). Therefore s = − ln(1−1/3) = ln 3−ln 2.

3. s(x) = ∑∞

n=1 x2n+1

n(2n+1) on x ∈ [−1, 1], and s =

∑∞ n=1

1 n(2n+1)

.

ANS: One can verify that the radius of convergence is R = 1. At x = 1, the series is s =

∑∞ n=1

1 n(2n+1)

, which is convergent, by comparing with 2- series. This also implies the convergence at x = −1. Hence, the interval of convergence is [−1, 1]. In particular, by Abel’s continuity theorem, one gets s =

∑∞ n=1

1 n3n

= s(1).

In the interval (−1, 1), one can have

s1(x) = s ′(x) =

∞ ∑

n=1

x2n

n .

By the previous problem, one knows that, by letting y = x2,

s1(x) = − ln(1 − x2) = − ln(1 + x) − ln(1 − x).

5

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Math 3001, Winter 2020 Homework 9

Notice that ∫

ln t dt = t ln t − ∫

t/t dt = t ln t − t.

Therefore,

s(x) = −[(1+x) ln(1+x)−(1+x)+(1−x)−(1−x) ln(1−x)] = 2x+(1−x) ln(1−x)−(1+x) ln(1+x).

So s(1) = limx→1− = 2 − 2 ln 2.

4. s = ∑∞

n=1 n2+1 n!2n

.

ANS: First of all, one considers the series

s(x) = ∞ ∑

n=1

(n2 + 1)xn

n! =

∞ ∑

n=1

nxn

(n − 1)! +

∞ ∑

n=1

xn

n! = s1(x) + s2(x).

One can verify the radius of convergence is R = ∞, and hence it is convergent in (−∞, ∞). Moreover, s = s(1/2). Let us deal with s2(x) =

∑∞ n=1 x

n/(n!). By calculating its derivative, one gets

(s2(x)) ′ = s2(x) − 1

which implies that s2(x) = e x − 1. (Or by the Taylor extension of ex =

∑∞ n=0 x

n/(n!), one gets s2(x) = e x − 1).

For s1(x) = ∑∞

n=1 nxn

(n−1)! , one has

s1(x) = ∞ ∑

n=1

nxn

(n − 1)! = x

∞ ∑

n=1

nxn−1

(n − 1)! .

Taking integral, one gets

s2(x) =

∫ x

0

s1(t)

t dt =

∞ ∑

n=1

xn

(n − 1)! = xex.

Hence s1(x) = (x 2 + x)ex.

In conclusion, one gets s(x) = s1(x) + s2(x) = (x 2 + x + 1)ex − 1. Hence

s = s(1/2) = 7e 1/2

4 − 1.

6

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