Math 2260 HW #12 Solutions

Math 2260 Written HW #12 Solutions

1. Does the series

(ln(n))2 n3

n=1

converge or diverge? Explain your answer.

Answer: Since (ln(n))2 goes to infinity as n , but grows relatively slowly, it's reasonable

to guess that it grows slower than n. Therefore, this series looks like it should be smaller

than the series

n 1

n3 =

n2 ,

n=1

n=1

which converges. Rather than messing around with a direct comparison (which would work), I'll use the limit comparison test, which means I should evaluate the limit

lim

n

(ln(n))2

n3 1 n2

(ln(n))2 n2

= lim

n

n3

? 1

(ln(n))2

= lim

.

n n

To evaluate this limit, we can take the corresponding limit of functions of x using L'H^opital's

Rule:

lim

(ln(x))2

=

lim

2

ln(x)

?

1 x

=

lim

2 ln(x) .

x x

x

1

x x

Using L'H^opital's Rule again yields

lim

2

?

1 x

=

lim

2 = 0.

x 1

x x

Therefore,

and, since the series also converges.

(ln(n))2

lim

n

n3

1 n2

=0

1 n2

converges,

the

Limit

Comparison

Test

tells

us

that

the

given

series

Notice that, since the limit was zero, this would have told us nothing if the series we were comparing to had diverged.

2. For each positive integer n, let

n

an =

2n 1

2n

if n is a prime number otherwise.

Does the series

n=1

an

converge

or

diverge?

Explain

your

answer.

Answer: I plan to use the Root Test. Notice that

lim

n

n

n 2n

= lim

n

nn =

2

1 ,

2

1

 since limn n n = 1, as we discussed in class. Likewise,

lim

n

n

1 2n

1 = lim

n 2

=

1 .

2

Thus,

no

matter

if

n

is

prime

or

not,

we

see

that

n an

is

approaching

1 2

as

n

;

in

other

words,

1

lim

n

n an

=

. 2

Since

1 2

<

1,

the

Root

Test

tells

us

that

the

series

converges.

3. Does the series

(-1)n+1

1

n ln(n)

n=2

converge absolutely, converge conditionally, or diverge? Explain your answer.

Answer:

First,

notice

that

the

terms

1 n ln(n)

are

positive,

decreasing,

and

approaching

zero,

so the hypotheses of the Alternating Series Test are satisfied, so we know that the series

converges. Therefore, we just need to determine whether or not it converges absolutely,

which is to say, whether or not the series

converges.

(-1)n+1

1

=

1

n ln(n)

n ln(n)

n=2

n=2

But now the thing to notice is that the terms look like a function that's easy to integrate, so the Integral Test will come in handy:

1

b1

dx = lim

dx.

2 x ln(x)

b 2 x ln x

Let

u

=

ln(x).

Then

du

=

1 x

dx

and,

since

u(2)

=

ln(2)

and

u(b)

=

ln(b),

the

above

integral

can be written as

ln(b) du

ln(b)

lim

= lim ln(u) = lim (ln(ln(b)) - ln(ln(2)))

b ln(2) u b

ln(2) b

Although it gets big incredibly slowly, limb ln(ln(b)) = , so the above integral diverges. (To emphasize just how incredibly slowly ln(ln(x)) goes to infinity, look at the below graph of the function. Even for x = 10100 [a.k.a. a googol], it's still smaller than 5.5!)

5.44

5.42

5.40

5.38

2 1099

4 1099

2

6 1099

8 1099

1 10100

Therefore, since the integral diverges, the Integral Test implies that the series

1

n=2 n ln(n)

diverges as well, which means that the original series only converges conditionally.

3

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