Math 115 HW #4 Solutions

Math 115 HW #4 Solutions

From ?12.5

8. Does the series converge or diverge?

(-1)n

n

n=1

n3 + 2

Answer: This is an alternating series, so we need to check that the terms satisfy the hy-

potheses of the Alternating Series Test. To see that the terms are decreasing in absolute value

(i.e. that bn+1 < bn), define

x

f (x) =

.

x3 + 2

Then

f (x) =

x3

+

2

-

x 2 x3+2

x2

? 3x2

=

2(x3 + 2) - 3x3

2x2 x3 + 2

=

4 - x3

.

2x2 x3 + 2

So long as x > 0, the term on the right is negative, so we see that f is a decreasing function. Therefore, the terms of the sequence are decreasing in absolute value.

To see that the terms are going to zero, we need to show that

n

lim

= 0.

n n3 + 2

In

the

lefthand

side,

multiply

both

numerator

and

denominator

by

1 n

.

This

yields

1

lim

n

1 n

n3

+2

=

lim

n

1 .

n

+

2 n2

Since the numerator is constant and the denominator goes to infinity as n , this limit is equal to zero.

Therefore, we can apply the Alternating Series Test, which says that the series converges.

12. Does the series converge or diverge?

(-1)n-1 e1/n n

n=1

Answer: Again, we want to use the Alternating Series Test, so we need to confirm that the terms are decreasing in absolute value and going to zero.

To see that the terms are decreasing, we want to show that

1

1

e n+1

en

1.

To see that the terms go to zero, consider the limit

1

lim

n

np

.

This limit is certainly zero since the numerator is constant and the denominator is going to

(because p > 0).

Therefore, the Alternating Series Test tells us that

(-1)n-1 np

converges

for

p > 0.

From ?12.6

12. Is the series

sin 4n

4n

n=1

absolutely convergent, conditionally convergent, or divergent?

Answer: Using the fact that | sin x| 1 for any x, we know that

sin 4n | sin 4n| 1 4n = |4n| 4n .

Since the series

1 4n

n=1

converges (since it's a geometric series), we know, by the Comparison Test, that

sin 4n 4n

n=1

converges. Thus, the series

n=1

sin 4n 4n

converges

absolutely.

18. Is the series

n!

nn

n=1

absolutely convergent, conditionally convergent, or divergent?

4

Answer: Using the Ratio Test,

lim

n

(n+1)! (n+1)n+1

n! nn

nn(n + 1)!

nn(n + 1)

=

lim

n

(n +

1)n+1n!

=

lim

n

(n + 1)n+1 .

Canceling a factor of n + 1 from both numerator and denominator yields

nn

lim

n

(n

+

1)n

.

Dividing numerator and denominator by nn gives

lim

n

1 nn

nn

1 nn

(n

+

1)n

=

lim

n

1

1

+

1 n

1 n=e

since limn

1

+

1 n

n

=

e.

Therefore,

since

1 e

<

1,

the

Ratio

Test

says

that

the

series

converges absolutely.

22. Is the series

-2n 5n

n+1

n=2

absolutely convergent, conditionally convergent, or divergent?

Answer: Using the Root Test,

lim n

n

-2n

5n

= lim n

2n 5n = lim

2n 5 .

n+1

n n + 1

n n + 1

Since

limn

2n n+1

= 2,

the

above

limit

is

equal

to

32,

which

is

certainly

>

1.

Therefore,

by

the Root Test, the series diverges.

24. Is the series

n

(ln n)n

n=2

absolutely convergent, conditionally convergent, or divergent?

Answer: Using the Root Test,

lim n

n

n (ln n)n

nn = lim = 0

n ln n

since limn n n = 1 and limn ln n = . Therefore, the Root Test says that the series converges absolutely.

30. The series

an is defined by the equations

a1 = 1

2 + cos n

an+1 =

n

an.

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download