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.

1

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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 Downloaded by Good Morning ([email protected]) lOMoARcPSD|8114622 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 Downloaded by Good Morning ([email protected]) lOMoARcPSD|8114622 https://www.studocu.com/en-ca?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=hw9solution-assignment-solutions 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. 4 Downloaded by Good Morning ([email protected]) lOMoARcPSD|8114622 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 Downloaded by Good Morning ([email protected]) lOMoARcPSD|8114622 https://www.studocu.com/en-ca?utm_campaign=shared-document&utm_source=studocu-document&utm_medium=social_sharing&utm_content=hw9solution-assignment-solutions 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 Downloaded by Good Morning ([email protected]) lOMoARcPSD|8114622