Math Real Analysis

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HW5 solution

Real Analysis II (Memorial University of Newfoundland)

StuDocu is not sponsored or endorsed by any college or university

HW5 solution

Real Analysis II (Memorial University of Newfoundland)

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Math 3001, Winter 2020 Homework 5

(a): Determine the convergence/divergence of the following series.

1. ∑∞

n=1 (−1)n

√ n

n+100 ;

ANS: First, consider the function

f(x) = x + 100/x, x > 0.

One can calculate its derivative and get

f ′(x) = 1 − 100/x2.

Notice that f ′(x) > 0 if and only if x2 > 100, i.e., x > 10. Hence f(x) is monotone increasing on x ∈ (10, ∞). Notice that √

n

n + 100 = 1/f(

√ n)

and is monotone decreasing on n ≥ 100 to 0. The series considered is an alternating series and by Lebnitz test, one gets the series is convergent. Moreover, one can compare with 1/2-series to get that

∑

√ n

n + 100

is divergent. Hence the series considered is conditionally convergent.

2. ∑∞

n=1 (−1)n cos(2n)

n ;

ANS: Let an = 1/n and one has an monotone decreasing to 0. Let bn = (−1)n cos(2n). We need to bound the partial sum of bn, i.e.,

Bn = − cos(2) + cos(4) − cos(6) + cos(8) + · · · + (−1)n cos(2n).

Multiply 2 cos(1) from both sides, one gets, by the formula 2 cos(x) cos(y) = cos(x+y)+cos(x) cos(y),

2 cos(1)Bn = −[cos(3) + cos(1)] + [cos(3) + cos(5)] + · · · + (−1)n[cos(2n − 1) + cos(2n + 1)] = (−1)n cos(2n + 1) − cos(1).

This further implies that

|Bn| = |(−1)n cos(2n + 1) − cos(1)|

2| cos(1)| ≤ 1/| cos(1)|

and hence Bn is bounded. Therefore, by Dirichlet test, one gets the series considered convergent.

3. ∑∞

n=1 (−1)n sin2(n)

n ; (Hint: you may need to use Problem a.2)

ANS: Here, we will use the following formula

2 sin2(n) = 1 − cos(2n).

Hence (−1)n sin2(n)

n =

(−1)n(1 − cos(2n)) 2n

.

We know that ∑

(−1)n/n is convergent. By the previous problem, one gets the series considered also converges.

1

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Math 3001, Winter 2020 Homework 5

4. ∑∞

n=1(−1)n ln n n ;

ANS: Let f(x) = ln x/x over x > 1. Calculating its derivative, one gets f ′(x) = (1 − ln x)/x2. Hence f ′(x) < 0 if x > e. Moreover, one can calculate the following limit by L-Hopital:

lim x→∞

ln x

x = lim

x→∞

1

x = 0.

Hence, after n > e, ln n n

is decreasing to 0. By Lebnitz, one gets the convergence of the series considered.

(b): Show that the Cauchy product of the series ∑∞

n=0 (−1)n √ n

with itself is divergent.

Proof. First of all, one can write the general terms for the Cauchy product as

cn = (−1)n n−1 ∑

k=1

1 √

k(n − k) .

To prove the divergence of ∑

cn, let us verify limn→∞ cn is not 0. We will use the following inequality: √

k(n − k) ≤ n/2 for all k = 1, · · · , n − 1. Hence,

|cn| ≥ (n − 1) × 2/n > 1, ∀n ≥ 3.

This implies that |cn| can not be approaching to 0 (and the gap is at least 1). Hence cn does not approach to 0 neither. As a consequence, one has the divergence of

∑

cn.

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