Math Real Analysis

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HW2 Solution

Real Analysis II (Memorial University of Newfoundland)

StuDocu is not sponsored or endorsed by any college or university

HW2 Solution

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 2

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

(a) Let |q| < 1 and x ∈ R, ∑∞

n=1 q n sin(nx);

ANS: This series is convergent. In fact, as |q| < 1, one gets that the geometric series

∑∞ n=1 |q|

n is convergent. Therefore, by Cauchy-Criterion, for all ǫ > 0, there is N(ǫ) > 0, such that, for all m > n > N(ǫ), one has

|q|n+1 + · · · + |q|m < ǫ.

For all x and n, one knows that | sin(nx)| ≤ 1. Therefore, for all ǫ > 0, and for all m > n > N(ǫ), one has

|qn+1 sin(nx + x) + · · · + qm sin(mx)| ≤ |q|n+1 + · · · + |q|m < ǫ.

Hence by Cauchy-Criterion, one gets the convergence of ∑∞

n=1 q n sin(nx).

(b) ∑∞

n=1( 2

n3 + 4−n)

ANS: This series is the sum of ∑∞

n=1 2

n3 and

∑∞ n=1 4

−n. The first one is twice of the p-series with p = 3 and hence convergent. The second one is a geometric series with ratio 1/4 which is also convergent. Hence the desired series is convergent.

(c) ∑∞

n=1(2 n + 3−n);

ANS: This series is the sum of ∑∞

n=1 2 n and

∑∞ n=1 3

−n. The first one is geometric series with ratio 2 and hence divergent. The second one is a geometric series with ratio 1/3 which is convergent. Hence the desired series is divergent.

(d) ∑∞

n=1( 2√ n + 3n).

ANS: This series is the sum of ∑∞

n=1 2

n1/2 and

∑∞ n=1 3

n. The first one is twice of the p-series with p = 1/2 and hence divergent. The second one is a geometric series with ratio 3 which is also divergent. As both of the series contain positive terms, hence the partial sum of the desired series is bigger than both of the partial sum of the two series which diverges to ∞. Hence the desired series is also divergent.

2. Let the series ∑∞

n=1 a 2 n and

∑∞ n=1 b

2 n be both convergent. Moreover, assume

that an, bn ≥ 0 for all n ≥ 1. Prove that the following series are convergent:

(a) ∑∞

n=1(an + bn) 2;

Proof. For all n ≥ 1, one knows that

(an + bn) 2 ≤ 2(a2n + b

2

n).

This can be seen from 2anbn ≤ a 2 n + b

2 n. Let sn be the partial sum of∑∞

n=1(an + bn) 2. One knows that, as each (an + bn)

2 ≥ 0, sn is monotone increasing. Moreover,

sn ≤ 2[ n∑

i=1

a2i + n∑

i=1

b2i ] ≤ 2[ ∞∑

n=1

a2n + ∞∑

n=1

b2n] = c.

Note that c is a finite number by assumption. Therefore sn is monotone increasing and bounded from above by a finite number, which implies the convergence of

∑∞ n=1(an + bn)

2.

(b) ∑∞

n=1 an n .

Proof. For all n ≥ 1, one knows that an n

≤ a2n+1/n

2

2 . Let sn be the

partial sum of ∑∞

n=1 an n . One knows that, as each an

n ≥ 0, sn is monotone

increasing. Moreover,

sn ≤ 1/2[ n∑

i=1

a2i + n∑

i=1

1/n2] ≤ 1/2 ∞∑

n=1

a2n + 1/2 ∞∑

n=1

1/n2 = c.

Note that c is a finite number by assumption. Therefore sn is monotone increasing and bounded from above by a finite number, which implies the convergence of

∑∞ n=1

an n .

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Math 3001, Winter 2020 Homework 2

3. Suppose that limn→∞ an = 0 and ∑∞

k=1(a2k + a2k−1) is convergent. Prove that∑∞ n=1 an is convergent.

Proof. Let sn be the partial sum of ∑∞

n=1 an. Let Bk be the partial sum of

∑∞ k=1(a2k + a2k−1), that is Bk = a1 + a2 + · · · + a2k−1 + a2k = s2k, for all

k = 1, 2, · · · . By assumption, one gets that Bk → B with some constant B. Consequently, s2k → B as well. This further implies that s2k+1 = s2k +a2k+1 → B. So we have prove that both odd and even subsequence of sn converge to the same number B, which further implies that sn → B. Thus, the desired series∑∞

n=1 an is convergent.

4. Suppose that the series ∑∞

n=1 an is convergent with an > 0 and an being mono- tone decreasing. Prove that

lim n→∞

nan = 0.

Proof. Let bn = nan be the desired sequence. By Cauchy-Criterion, for all ǫ > 0, there is N(ǫ) > 0, such that, for all m > n > N(ǫ), one has

an+1 + · · · + am < ǫ.

Note that an ≥ an+1 for all n. Let m = 2n with n > N(ǫ), one gets

na2n ≤ an+1 + · · · + am < ǫ.

That is, we have prove that, for all ǫ > 0, there is N(ǫ) > 0, such that, for all n > N(ǫ),

2na2n < ǫ.

This leads to b2n → 0 as n → ∞.

Similarly, let m = 2n + 1 with n > N(ǫ), one gets

(n + 1)a2n+1 ≤ an+1 + · · · + am < ǫ.

That is, we have prove that, for all ǫ > 0, there is N(ǫ) > 0, such that, for all n > N(ǫ),

(n + 1)a2n+1 < ǫ.

This leads to (n + 1)a2n+1 → 0 as n → ∞. Note that the convergence of∑∞ n=1 an implies that an → 0. Therefore,

b2n+1 = (2n + 1)a2n+1 = 2(n + 1)a2n+1 − a2n+1 → 0.

In summary, we have proved that both odd and even subsequence of bn con- verges to 0. Hence bn = nan → 0 as desired.

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