Math Real Analysis

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HW3 Solution

Real Analysis II (Memorial University of Newfoundland)

StuDocu is not sponsored or endorsed by any college or university

HW3 Solution

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 3 Due: Feb 10, 2020

1. Determine the convergence/divergence of the following series.

(a) ∑

∞

n=1 sin(1/n 2);

ANS: Compare with 2-series as follows:

lim x→0

sin(x)

x = lim

x→0 cos(x) = 1.

Therefore,

lim n→∞

sin(1/n2)

1/n2 = 1

and hence the desired series is convergent by the convergence of 2-series.

(b) ∑

∞

n=3 1 n tan(π/n);

ANS: Compare with 2-series as follows:

lim n→∞

1 n tan(π/n)

1/n2 = lim

n→∞

tan(π/n)

1/n = lim

x→0

tan(πx)

x = lim

x→0 π sec(πx)2 = π

and hence the desired series is convergent by the convergence of 2-series.

(c) ∑

∞

n=1

(

ln n n

)3 ;

ANS: Compare this series with ∑

∞

n=1 1

n3/2 as follows:

L = lim n→∞

n3/2 ln3 n

n3 = lim

n→∞

ln3 n

n3/2 = 0.

Hence the desired series is also convergent as ∑

∞

n=1 1

n3/2 is convergent.

(d) ∑

∞

n=1 ln(1 + 1/n); ANS: Compare with 1-series as follows:

lim x→0

ln(1 + x)

x = lim

x→0

1

1 + x = 1.

Therefore,

lim n→∞

ln(1 + 1/n)

1/n = 1

and hence the desired series is divergent by the divergence of 1-series.

(e) ∑

∞

n=1 n( √ n4 + 1 −

√ n4 − 1);

ANS: Note that

lim n→∞

n( √ n4 + 1 −

√ n4 − 1)

1/n = lim

n→∞

2n2√ n4 + 1 +

√ n4 − 1

= 1.

Note that ∑

∞

n=1 1/n is divergent and hence the series ∑

∞

n=1 n( √ n4 + 1 −√

n4 − 1) is also divergent.

1

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Math 3001, Winter 2020 Homework 3 Due: Feb 10, 2020

(f) ∑

∞

n=1 1

2n2[2+sin(nπ)] ;

ANS: Note that 0 ≤ 1 2n2[2+sin(nπ)]

≤ 1 2n2

and the series ∑

1 2n2

is convergent. By comparison test, the desired series is also convergent.

(g) ∑

∞

n=1 n+2 sin(nx) n2+sin(nx)

, x ∈ [0, 2π]; ANS: Note that

0 ≤ n − 2 n2 + 1

≤ n + 2 sin(nx)

n2 + sin(nx) ≤

n + 2

n2 − 1 ; n ≥ 3.

By comparing with ∑

1/n, we know the series ∑

n−2 n2+1

is divergent. By comparison test, the desired series is also divergent.

(h) ∑

∞

n=1 1

n1−1/n ;

ANS: We compare the desired series with ∑

1/n as follows:

L = lim n→∞

n

n1−1/n = lim

n→∞ n1/n = 1

and hence the desired series is divergent.

(i) ∑

∞

n=1 2+sin2(n)

3n .

ANS: Note that

0 ≤ 2 + sin2(n)

3n ≤

3

3n =

1

3n−1 .

The right series is a geometric series with ratio 1/3 and hence it is con- vergent. By comparison test, the desired series is also convergent.

2. Use both Cauchy-Condensation test and Integral test for the convergence of the following series.

(a) ∑

∞

n=4 1

n(ln n)(ln ln n) ;

ANS: Use the Cauchy condensation test, the convergence of the desired series is equivalent to the convergence of the series

∞ ∑

n=4

2n

2n(ln 2n)(ln ln 2n) =

1

ln 2

∞ ∑

n=4

1

n(ln ln 2 + ln n) ,

which is further equivalent to the convergence of the following series, by Cauchy Condensation test again,

1

ln 2

∞ ∑

n=4

2n

2n(ln ln 2 + ln 2n) =

1

ln 2

∞ ∑

n=4

1

(ln ln 2 + ln 2 · n) .

2

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Math 3001, Winter 2020 Homework 3 Due: Feb 10, 2020

The last series is divergent by comparing with ∑

1/n. Hence the desired series is also divergent.

Now for integral test, let f(x) = 1 x(ln x)(ln ln x)

, which is clearly monotone decreasing to 0 as x increases to ∞. Therefore, the convergence of the desired series is equivalent to

∫

∞

x=4

1

x(ln x)(ln ln x) dx < ∞.

Here, by letting u = ln ln x, one has

∫

∞

x=4

1

x(ln x)(ln ln x) dx = ln ln ln x|∞x=4 = ∞.

Hence, the desired series is divergent.

(b) ∑

∞

n=2 1

n(ln n)p , p > 0

ANS: First of all, one can check that an = 1

n(ln n)p is monotone decreasing

to 0 if p > 0. Therefore, the convergence of

∞ ∑

n=2

1

n(ln n)p , p > 0

is equivalent to that of

∞ ∑

n=2

1

(ln 2n)p =

∞ ∑

n=2

1

(ln 2)pnp , p > 0.

By the convergence of p-series, if p > 1, one gets

∞ ∑

n=2

1

n(ln n)p

is convergent; and if 0 < p ≤ 1, one gets ∞ ∑

n=2

1

n(ln n)p

is divergent.

3

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Math 3001, Winter 2020 Homework 3 Due: Feb 10, 2020

Now for integral test, one let f(x) = 1 x(ln x)p

, and one can check that

f(x) ≥ 0 is monotone decreasing to 0 as x increases to ∞. Therefore, the convergence of

∞ ∑

n=2

1

n(ln n)p , p > 0

is equivalent to ∫

∞

x=2

1

x(ln x)p dx, p > 0.

Here ∫

∞

x=2

1

x(ln x)p =

∫

∞

ln 2

1

yp dy = lim

A→∞

1

1 − p (A1−p − (ln 2)1−p)

which is divergent if 0 < p ≤ 1 and convergent if p > 1.

3. Prove that if ∑

∞

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

∞

n=1 a 2 n is also

convergent. Proof. Note that the series

∑

∞

n=1 an convergent implies that limn→∞ an = 0. This further implies that 0 ≤ an ≤ M for some M < ∞ and for all n ≥ 1. Hence, one has

a2n = an · an ≤ Man, n ≥ 1. By comparison test,

∑

∞

n=1 a 2 n is also convergent (Note that, you can not use

the limit form, because an may be 0.)

4

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