−nx2[1 − 2nx2] = 0 and hence x2 = 1/(2n). However, if n is big enough, then x2 = 1/(2n) will be out of the interval [1, 2]. Moreover, if x ∈ [1, 2], fn(x) will be decreasing for all n big enough. Hence, |fn(x)| ≤ ne−n. Let Mn = n/e

n, one can verify that lim Mn = 0 as follows

lim n→∞

en = lim

x→∞

ex = lim

x→∞ 1/ex = 0.

By M-test for the uniform convergence of sequence, one gets fn unif conv to 0. Since each fn is continuous on [1, 2], and hence they are Riemann integrable. By integration term-by-term, one gets the limit is 0.

2. limn→∞ ∫ 1 0 (1 + x/n)n dx

ANS: First of all, fn(x) ≤ (1 + 1/n)n

for all x ∈ [0, 1]. Moreover, (1 + 1/n)n is monotone increasing to e. Hence, |fn(x)| ≤ e for all x ∈ [0, 1] and n ≥ 1. Let M = e which is Riemann integrable on x ∈ [0, 1]. One can also check that fn(x) → ex as n → ∞. Each fn and ex are Riemman integrable on x ∈ [0, 1]. Hence, by the bounded convergence theorem, one gets the desired limit is

∫ 1 0 ex dx = e − 1.

(b): Prove ∑

fn(x) is not uniformly convergent on the set D, by checking fn(x) does not uniform converge to 0 on the set D.

1. ∑∞

n=1[n(x 1/n − 1) − ln x], on D = [a, ∞) with a > 0.

ANS: First of all, one can calculate

lim n→∞

n(x1/n − 1) = lim y→0

(xy − 1) y

= ln x

by letting y = 1/n. Hence fn(x) = [n(x 1/n − 1) − ln x] → 0 as n → ∞ pointwise. Now let us pick

up xn = e n, and one can check that fn(e

n) = n(e − 2) → ∞. Hence fn is not uniformly convergent to 0, and the series is not uniform convergent.

2. ∑∞

n=1 ln(1+nx2)

2n , on D = [0, ∞).

ANS: First of all, one can calculate

lim n→∞

ln(1 + nx2)

2n = lim

y→∞

ln(1 + yx2)

2y = lim

y→∞

2(1 + yx2) = 0.

Hence fn(x) → 0 as n → ∞ pointwise. Now let us pick up xn = en, and one can check that fn(e

n) ≥ 1, and hence supx∈[0,∞) fn is not approaching to 0. That is, fn is not uniformly convergent to 0, and the series is not uniform convergent.

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Math 3001, Winter 2020 Homework 8

3. ∑∞

n=1 xn

1+x2n , on D = [0, 1).

ANS: First of all, we pick up x1n = 1 − 1/n and x2n = 1 − 2/n. Both sequence go to 1. However, one can calculate that fn(x

1 n) → e

−1

1+e−2 , while fn(x

2 n) → e

−2

1+e−4 . These two limits are not equal to each

other. That is, fn is not uniformly convergent to 0, and the series is not uniform convergent.

4. ∑∞

n=1 n2x

1+n3x2 , on D = (0, ∞).

ANS: First of all, we pick up x1n = 1/n 2 and x2n = 2/n

2. Both sequences go to 0, however, one can calculate that fn(x

1 n) → 1, while fn(x2n) → 2. These two limits are not equal to each other. That

is, fn is not uniformly convergent to 0, and the series is not uniform convergent.

(c): Use M-test to check fn → 0 uniformly on D, and also ∑

fn absolutely uniform converges on D.

1. fn(x) = 2nx

enx 2 , on D = [a, ∞) with a > 0.

ANS: First of all, f ′n = 2ne

−nx2[1 − 2nx2] = 0 and hence x2 = 1/(2n). However, if n is big enough, then x2 = 1/(2n) will be out of the interval [a, ∞) for a > 0. Moreover, if x ∈ [a, ∞), fn(x) will be decreasing for all n big enough. Hence, |fn(x)| ≤ 2nae−na. Let Mn = 2na/ena, one can verify that lim Mn = 0 as follows

lim n→∞

ena = lim

x→∞

ex = lim

x→∞ 1/ex = 0.

By M-test for the uniform convergence of sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, by ratio test as follows:

r = lim n→∞

n + 1

ena+a ×

ena

n = 1/ea < 1.

Hence, the M-test implies that ∑

fn(x) is also uniform convergent in [a, ∞).

2. fn(x) = 1

nenx , on D = [a, ∞) with a > 0.

ANS: First of all, for each n, one notes that fn(x) is decreasing on [a, ∞) as enx is increasing. Hence, |fn(x)| ≤ 1nena . Let Mn =

1 nena

, one can verify that lim Mn = 0 as e na increasing to ∞. By

M-test for the uniform convergence of sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, by ratio test as follows:

r = lim n→∞

nena

(n + 1)ena+a = 1/ea < 1.

Hence, the M-test implies that ∑

fn(x) is also uniform convergent in [a, ∞).

3. fn(x) = xe −n2x, on D = [0, ∞)

ANS: First of all, f ′n = e

−n2x[1 − n2x] = 0 and hence x = 1/n2. If x > 1/n2, f ′n < 0 and if x < 1/n

2, f ′n > 0. Therefore, fn attains its maximal at x = 1/n2, and |fn| ≤ 1en2 . Let Mn =

1 en2

which approaches to 0. By M-test for the uniform convergence of sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, as it is essentially the 2-series up to a constant. Hence, the M-test implies that

∑

fn(x) is also uniform convergent in [a, ∞).

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Math 3001, Winter 2020 Homework 8

4. fn(x) = 2x2

1+n3x4 , on D = (−∞, ∞).

ANS: First of all, recall that a2 + b2 ≥ 2|ab|. For each n, one notes that |fn(x)| ≤ 2x2/(2n3/2x2), x 6= 0

and fn(0) = 0. Hence, |fn(x)| ≤ 1/n3/2. Let Mn = 1/n3/2, one can verify that lim Mn = 0 By M-test for the uniform convergence of sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, as it is a 3/2-series. Hence, the M-test implies that ∑

fn(x) is also uniform convergent in [a, ∞).

5. fn(x) = (

ln x x

)n , on D = [1, ∞).

ANS: First of all, the maximal of ln x/x is obtained at x = e. This follows form (

ln x

)′

= 1 − ln x

= 0 only if x = e, and > 0 if x < e, and < 0 if x > e. Hence ln x/x ≤ 1/e and |fn| < e−n = Mn. One can verify that lim Mn = 0.By M-test for the uniform convergence of sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, as it is a geometric series with ratio 1/e < 1. Hence, the M-test implies that

∑

fn(x) is also uniform convergent in [a, ∞).

6. fn(x) = sin(nx) n √ n

, on D = (−∞, ∞).

ANS: First of all,

|fn(x)| = | sin(nx)

n √ n

| ≤ 1/n3/2, ∀x ∈ (−∞, ∞).

Let Mn = 1/n 3/2, one can verify that lim Mn = 0. By M-test for the uniform convergence of

sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, as it is a 3/2-series. Hence, the M-test implies that ∑

fn(x) is also uniform convergent in [a, ∞).

7. fn(x) = x2

(1+x2)n , on D = [a, ∞) with a > 0.

ANS: First of all,

|fn(x)| = | x2

(1 + x2)n | ≤ 1/(1 + x2)n−1 ≤

(1 + a2)n−1 , ∀x ∈ (−∞, ∞).

Let Mn = 1

(1+a2)n−1 , one can verify that lim Mn = 0. By M-test for the uniform convergence of

sequence, one gets fn unif conv to 0.

Note that ∑

Mn is also convergent, as it is a geometric series with ratio r = 1

(1+a2) < 1. Hence, the

M-test implies that ∑

fn(x) is also uniform convergent in [a, ∞). (d): Use Dirichlet and/or Abel test to prove the uniform convergence of the following series.

1. ∑∞

n=1 (−1)n

ln n+x2 , on D = [a, b] with finite a, b.

ANS: Let an = (−1)n. Then |An| = |a1 +· · ·+an| ≤ 1, which implies that An is uniformly bounded for all n and all x. Let bn(x) =

1 ln n+x2

. For each x given, one can check that bn(x) ≥ bn+1(x). Moreover, |bn(x)| ≤ 1/ ln n → 0. By M-test, bn is uniformly convergent to 0. Hence, one gets the uniform convergence of the series, by Dirichlet-test.

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Math 3001, Winter 2020 Homework 8

2. ∑∞

n=1(−1)n sin(x/n) is uniformly convergent on any finite interval [a, b].

ANS: Let an = (−1)n. Then |An| = |a1 + · · · + an| ≤ 1, which implies that An is uniformly bounded for all n and all x. Let bn(x) = sin(x/n). For each a, b given, there exists a N0, such that x/n ∈ [−π/2, π/2] for all x ∈ [a, b]. Hence, sin(x/n) is monotone increasing in the interval [a, b] for all n ≥ N0. Moreover, |bn(x)| ≤ max{| sin(a/n)|, | sin(b/n)|} → 0. By M-test, bn is uniformly convergent to 0. Hence, one gets the uniform convergence of the series, by Dirichlet-test.

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