Math 115 HW #3 Solutions

[Pages:6]Math 115 HW #3 Solutions

From ?12.2

20. Determine whether the geometric series

en 3n-1

n=1

is convergent or divergent. If it is convergent, find its sum. Answer: I can re-write the terms as

en

en-1

e n-1

3n-1 = e 3n-1 = e 3

.

Therefore, the series

en

e n-1

en

3n-1 = e

3

=e

, 3

n=1

n=1

n=0

where

the

second

equality

comes

from

shifting

the

index

by

one.

Since

e 3

<

1,

we

know

that

the geometric series

en

1

3

3

n=0

=

1-

e 3

=

. 3-e

Therefore, the given series converges and the sum is given by

en

en

3

3e

3n-1 = e

=e

=

.

3

3-e 3-e

n=1

n=0

24. Determine whether the series

k(k + 2) (k + 3)2

k=1

is convergent or divergent. If it is convergent, find its sum.

Answer: This series diverges. To see this, I will show that the terms in the sequence do not

go to zero:

k(k + 2)

k2 + 2k

lim

k

(k + 3)2

= lim

k

k2 + 6k + 9 .

Dividing numerator and denominator by k2 yields

lim

k

1 k2

1 k2

1

+

2 k

1

+

6 k

+

9 k2

=

lim

k

1

1 +

+

2 k

6 k

+

9 k2

= 1.

Therefore, using the nth term test (a.k.a. Test for Divergence), the series diverges.

1

38. Determine whether the series

n

ln

n+1

n=1

is convergent or divergent by expressing sn as a telescoping sum. If it is convergent, find its sum.

Answer: We can re-write the terms in the series as

n

ln

= ln(n) - ln(n + 1).

n+1

Therefore, the partial sum

sn = a1 + a2 + a3 + . . . + an = (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) . . . + (ln(n) - ln(n + 1)) = ln(1) + (- ln(2) + ln(2)) + (- ln(3) + ln(3)) + . . . + (- ln(n) + ln(n)) - ln(n + 1) = ln(1) - ln(n + 1) = - ln(n + 1).

Therefore,

n

ln

n+1

n=1

so the given series diverges.

=

lim

n

sn

=

lim (-

n

ln(n

+

1))

=

-,

48. Find the values of x for which the series

(x - 4)n

n=1

converges. Find the sum of the series for those values of x.

Answer: Notice that this is a geometric series, so the series converges when |x - 4| < 1, meaning that

3 < x < 5.

Moreover, for such values of x, the series converges to

1

1

=

.

1 - (x - 4) 4 - x

70. If an and bn are both divergent, is (an + bn) necessarily divergent? Answer: No. Let an = 1 for all n and let bn = -1. Then

an = 1 diverges

n=1

n=1

and

bn = (-1) diverges

n=1

n=1

2

However, certainly converges.

(an + bn) = (1 + (-1)) = 0 = 0

n=1

n=1

n=1

From ?12.3

16. Determine whether the series

n2 n3 + 1

n=1

is convergent or divergent.

Answer:

If

we

let

f (x)

=

x2 x3+1

,

then

the

terms

of

the

series

and

the

function

f

satisfy

the

hypotheses of the Integral Test, so the series will converge if and only if

x2

1

f (x)dx =

1

x3 + 1 dx

is finite. Letting u = x3 + 1, we have that du = 3x2dx, so I can re-write the above integral as

1 du 1

= ln |u| ,

3 u=2 u 3

2

which diverges since ln(u) as u . Therefore, the series diverges by the Integral Test.

22. Determine whether the series

1

n(ln n)2

n=2

is convergent or divergent.

Answer:

If

we

let

f (x)

=

1 x(ln x)2

,

then

the

terms

of

the

series

and

the

function

f

satisfy

the

hypotheses of the Integral Test, so the series will converge if and only if

1

2

f (x)dx =

2

x(ln x)2 dx

is finite.

Letting

u

=

ln x,

we

have

that

du

=

1 x

dx,

so

I

can

re-write

the

above

integral

as

du u=ln 2 u2 =

- u-1

1

=,

u=ln 2 ln 2

which is finite. Therefore, the series converges by the Integral Test.

3

30. Find the values of p for which the series

ln n np

n=1

is convergent.

Answer: When p 0 the terms in the series do not go to zero, so the series will diverge.

When

p

>

0,

the

function

f (x)

=

ln x xp

and

the

series

satisfy

the

hypotheses

of

the

Integral

Test, so the series will converge if and only if

ln x

f (x)dx =

1

1

xp dx

(*)

is finite.

When p = 1, I can write the above integral as

ln x dx.

1x

Letting

u

=

ln x,

we

have

du

=

1 x

dx,

so

this

is

equal

to

u2

udu =

,

u=0

20

which is infinite.

When p = 1, I will use integration by parts to evaluate the integral in (*). Letting u = ln x

and

dv

=

dx xp

,

I

have

that

u = ln x 1

du = dx x

dv = x-pdx

x1-p

v=

,

1-p

so the anti-derivative is given by

ln x

x1-p

1

1

xp

dx

=

(ln x) 1

-p

-

1-p

xp dx

x1-p

1 x1-p

= (ln x)

-

1-p 1-p1-p

=

x1-p

(1

- p) ln x (1 - p)2

-

1

.

Hence,

1

ln x xp dx

=

x1-p (1

- p) ln x - (1 - p)2

1

,

1

which is finite only when 1 - p < 0.

Therefore, the series converges when p > 1.

4

34. Find the sum of the series

n=1

1/n5

correct

to

three

decimal

places.

Answer: If we estimate the sum by the nth partial sum sn, then we know that the remainder

Rn is bounded by

1

1

n+1 x5 dx Rn n x5 dx.

This means that

1

11 1

Rn

n

x5 dx

=

- 4

x4

n

= 4n4 ,

so the estimate will be accurate to 3 decimal places when this expression is less than 0.001.

In other words, we want to know for what n is it true that

1

1

4n4

<

. 1000

Solving for n, we get that

n > 4 250 3.98.

So letting n = 4, we have that

1111

11

1

s4 = 15 + 25 + 35 + 45 = 1 + 32 + 243 + 1024 1.036

is an estimate of the sum of the series that is correct to three decimal places.

From ?12.4

18. Determine whether the series converges or diverges.

1 2n + 3

n=1

Answer: Use the Limit Comparison Test to compare this series to

1 n

.

We

see

that

lim

n

1 2n+3

1 n

n = lim

n 2n + 3

=

1 .

2

Therefore, since

1 n

diverges,

the

Limit

Comparison

Test

tells

us

that

the

series

1 2n+3

also

diverges.

26. Determine whether the series converges or diverges.

n+5

n=1 3 n7 + n2

Answer: Use the Limit Comparison Test to compare this series to

1 n4/3

.

We

see

that

lim

n

3 nn7++5n2

1 n4/3

(n + 5)n4/3 = lim

n 3 n7 + n2

n7/3 + 5n4/3

= lim

.

n 3 n7 + n2

5

Therefore, dividing both numerator and denominator by n7/3, we see that this limit is equal

to

lim

n

1 n7/3

n7/3 + 5n4/3

1 n7/3

3 n7 + n2

= lim

1

+

5 n

= 1.

n

3

1

+

1 n5

Therefore, since

1 n4/3

converges,

the

Limit

Comparison

Test

tells

us

that

the

given

series

also converges.

32. Determine whether the series converges or diverges.

1 n1+1/n

n=1

Answer: Use the Limit Comparison Test to compare this series to

1 n

.

We

see

that

lim

n

1 n1+1/n

1 n

n

=

lim

n

n1+1/n

n

=

lim

n

n

?

n1/n

= lim

n

1 nn

=1

since limn n n = 1. Therefore, since

1 n1+1/n

also

diverges.

1 n

diverges,

the

Limit

Comparison

test

tells

us

that

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