Math 2260 Exam #3 Practice Problem Solutions

Math 2260 Exam #3 Practice Problem Solutions

1. Does the following series converge or diverge? Explain your answer. 2n 3n + n3 .

n=0

Answer: Since 3n + n3 > 3n for all n 1, it follows that

2n

2n

2n

3n + n3 < 3n = 3 .

Therefore,

2n

2n

1

3n + n3 <

3

n=0

n=0

=

1-

2 3

= 3.

Hence, the given series converges.

2. Does the following series converge or diverge? Explain your answer.

n 3n .

n=1

Answer: Use the Ratio Test:

lim

n

n+1 3n+1

n 3n

n + 1 3n

= lim

n

3n+1

?

n

1 n+1 = lim ?

n 3 n

=

1 .

3

Since

1 3

<

1,

the

Ratio

Test

implies

that

this

series

converges.

3. Does the following series converge or diverge? Explain your answer.

1

2n sin

.

n

n=1

Answer: Notice that the terms of this series are not going to zero:

1

1

lim 2n sin

= lim 2x sin

n

n

x

x

=

lim

x

sin

1

1 x

2x

=

lim

x

cos

1 x

?

-2

(2x)2

-2 x2

-2 cos

= lim

x

x2

1 x

4x2 ?

-2

1

= lim 4 cos

x

x

=4

where I went from the second to the third lines using L'H^opital's Rule. Since the limit of the terms is equal to 4, not zero, the series must diverge.

1

4. Does the following series converge or diverge? If it converges, find the sum. If it diverges, explain why. 2n + 3n 4n .

n=1

Answer: Re-writing slightly, the given series is equal to

2n 3n 4n + 4n

n=1

2n 3n

=

4n +

4n .

n=1

n=1

Since both of these series are convergent geometric series, I know the original series converges, so it remains only to determine the sum. Notice that

2n 2 4 8

2

24

2 2 n-1

2/4

1/2

4n = 4 + 16 + 64 + . . . = 4

1+ + +... 4 16

=

44

=

= = 1.

1 - 2/4 1/2

n=1

n=1

Similarly,

3n 3 9 27

3

39

3 3 n-1

3/4

3/4

4n = 4 + 16 + 64 + . . . = 4

1+ + +... 4 16

=

44

=

= = 3.

1 - 3/4 1/4

n=1

n=1

Therefore,

2n + 3n 2n 3n

4n = 4n + 4n = 1 + 3 = 4.

n=1

n=1

n=1

5. Find the interval of convergence of the power series

(2x - 5)n n23n .

n=1

Answer: We use the Ratio Test on the series of absolute values to first determine the radius of convergence:

lim

n

(2x-5)n+1 (n+1)2 3n+1

(2x-5)n n2 3n

|2x - 5|n+1

n3n

|2x - 5| n2

|2x - 5|

=

lim

n

(n + 1)23n+1

?

|2x - 5|n

=

lim

n

3

? (n + 1)2 =

. 3

Therefore,

the

given

series

converges

absolutely

when

|2x-5| 3

<

1,

meaning

when

|2x - 5|

<

3.

Now we check the endpoints. When 2x - 5 = 3, the series becomes

3n

1

n23n =

n2 ,

n=1

n=1

which converges.

Likewise, when 2x - 5 = -3, then series becomes

(-3)n (-1)n3n (-1)n

n23n =

n23n =

n2 ,

n=1

n=1

n=1

which also converges.

Therefore, the series converges for all x so that

-3 2x - 5 3,

which is the interval [1, 4].

2

6. Use the first two non-zero terms of an appropriate Taylor series to approximate

1

sin(x2) dx.

0

Estimate the error of your approximation (i.e. the difference between your answer and the actual value of the integral).

Answer: First, recall that the Taylor series centered at x = 0 for sin(x) is

x3 x5 x7 sin(x) = x - + - + . . . .

3! 5! 7! Therefore, the Taylor series centered at x = 0 for sin(x2) is

sin(x2) = (x2) - (x2)3 + (x2)5 - (x2)7 + . . . = x2 - x6 + x10 - x14 + . . . .

3!

5!

7!

3! 5! 7!

Hence,

1

sin(x2) dx =

1

x2 - x6 + x10 - . . .

dx =

x3 x7 -

+

x11

1 11 -... = - +

1

-. . . .

0

0

3! 5!

3 7 ? 3! 11 ? 5!

0 3 42 11 ? 5!

Therefore, we can approximate this number by

1 1 13 - =,

3 42 42

and we know the error is no bigger than

1

1

1

=

=,

11 ? 5! 11 ? 120 1320

so

in

particular

the

estimate

13 42

is

accurate

to

3

decimal

places.

(It

would

be

totally

fine

in

an

exam

situation

to

leave

your

answer

as

1 11?5!

.)

7. Find the radius of convergence of the Taylor series

(-1)n n2

1

+

n

(x

-

2)n.

n=2

Answer: Use the Ratio Test on the series of absolute values:

lim

n

(-1)n+1 1+(1+n) (n+1)2

(x

-

2)n+1

(-1)n n2

1+n

(x

-

2)n

2

+

n

|x

-

2|n+1

n2

= lim

n

(n + 1)2

? 1 + n |x - 2|n

2 + n n2

=

lim

n

|x

-

2|

?

1

+

n

?

(n

+

1)2

= |x - 2|.

Therefore, the series converges absolutely when |x - 2| < 1, so the radius of convergence is equal to 1.

3

 8. Write the second-degree Taylor polynomial for f (x) = x centered at c = 25. Use this polynomial to

approximate 26 and estimate the error of this approximation.

Answer: I can write down the Taylor series centered at x = 25 for the function f (x) = x by first

computing the derivatives of f :

f (x) = 1 x-1/2 = 1

2

2x

f (x) = 1 ? -1 x-3/2 = -1

22

4x3/2

f (x) = -1 ? -3 x-5/2 = 3

42

8x5/2

Therefore,

f (25) = 25 = 5

f (25) = 1

1 =

2 25 10

-1

-1 -1

f

(25) =

4 ? 253/2

=

=

4 ? 125

500

3

3

3

f (25) =

=

=

8 ? 255/2 8 ? 3125 25, 000

Hence, the Taylor series centered at x = 25 for x is

(x - 25) (x - 25)2 3(x - 25)3

x=5+

-

+

-...

10

1000

150, 000

Therefore,

(26 - 25) (26 - 25)2

11

26 5 +

-

=5+ -

= 5.1 - 0.001 = 5.099,

10

1000

10 1000

with an error no more than

3(26 - 25)3

3

1

=

=

= 0.00002.

150, 000 150, 000 50, 000

(In fact, 26 5.0990195.)

9. Does the series

nn

n2

n=1

converge or diverge. Be sure to give a complete explanation.

Answer: Since limn n n = 1 as we discussed in class, a limit comparison to the series

natural:

lim

n

n n

n2 1 n2

nn

n2

= lim

n

n2

?

1

= lim n n = 1.

n

1 n2

is a

Therefore, since the series

1 n2

converges,

the Limit

Comparison Test

implies

that the

given

series

converges as well.

4

10. Does the following series converge or diverge? Explain your answer. n!(n + 1)! . (3n)!

n=1

Answer: Use the Ratio Test:

(n+1)!(n+2)!

lim

n

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

(n + 1)!(n + 2)! (3n)!

(n + 1)(n + 2)

= lim

?

= lim

= 0,

n (3n + 3)! n!(n + 1)! n (3n + 3)(3n + 2)(3n + 1)

(3n)!

since the numerator is a polynomial of degree 2 but the denominator is a polynomial of degree 3. Therefore, since 0 < 1 the Ratio Test implies that the series converges.

11. Does the sequence

n2

arctan n2 + 1 n=1

converge or diverge? If it converges, find the limit; if it diverges, explain why.

Answer: First, notice that

n2

lim

n

n2

+1

=

1.

Therefore, the term inside the arctangent is going to 1, so

n2

lim arctan

n

n2 + 1

= arctan(1) = . 4

12. Does the series

1

n2 - n

n=2

converge or diverge? Explain your answer.

Answer:

For

large

n,

the

n2

should

dominate

the

n,

so

let's

do

a

limit

comparison

to

the

convergent

series

1 n2

.

lim

n

1 n2- n

1 n2

1

n2

=

lim

n

n2

-

n

?

1

n2

1

=

lim

n

n2

-n

=

lim

n

1-

1 n3/2

= 1.

Therefore, since

1 n2

converges,

so

does

n2

1 -n

.

13. Does the series converge or diverge? Explain your answer.

n! n5

n=1

Answer: Use the Ratio Test:

(n+1)!

lim

n

(n+1)5 n!

n5

(n + 1)! n5

=

lim

n

(n

+ 1)5

?

n!

n5

=

lim (n

n

+

1)

(n

+

1)5

=

since

the

expression

n5 (n+1)5

is

going

to

1

but

(n + 1)

is

going

to

.

Therefore, the Ratio Test implies that the series diverges.

5

14. Does the series converge or diverge? Explain your answer.

3n n3

n=1

Answer: Use the Ratio Test:

lim

n

3n+1 (n+1)3

3n n3

3n+1 n3

=

lim

n

(n

+

1)3

?

3n

(n + 1)3

= lim 3 ?

n

n3

= 3.

Since 3 > 1, the Ratio Test implies that the series diverges.

15. Does the series

2n + 3

(n2 + 3n + 6)2

n=0

converge or diverge? Explain your answer.

Answer: For n very large, the denominator will be dominated by the term n4, so do a limit comparison

to the convergent series

n n4

:

lim

n

2n+3 (n2 +3n+6)2

n n4

2n + 3

n4

=

lim

n

(n2

+

3n

+

6)2

?

n

2n + 3

n4

= lim

n

n

? (n2 + 3n + 6)2

= 2 ? 1 = 2.

Therefore, since the limit is finite and the series

n n4

=

1 n3

converges,

the

Limit

Comparison

Test

implies that the given series converges as well.

16. For which values of x does the series

(x - 4)n 5n

n=0

converge? What is the sum of the series when it converges?

Answer: First, use the Ratio Test on the series of absolute values:

lim

n

(x-4)n+1 5n+1

(x-4)n 5n

|x - 4|n+1

5n

|x - 4|

= lim

n

5n+1

? |x - 4|n =

, 5

so

the

given

series

converges

absolutely

whenever

|x-4| 5

<

1,

meaning

when

|x - 4|

<

5

(from

this

we

see that the radius of convergence of the series is 5).

Now check the endpoints. When x - 4 = 5, the series becomes

5n

5n = 1,

n=0

n=0

which diverges. Similarly, when x - 4 = -5, the series becomes

(-5)n 5n

=

(-1)n5n 5n

=

(-1)n,

n=0

n=0

n=0

which also diverges.

6

Therefore, the series converges for

-5 < x - 4 < 5,

which is to say, on the interval (-1, 9).

When the series does converge, it is just the geometric series

x-4 n

1

1

5

5

n=0

=

1

-

x-4 5

=

5-(x-4) 5

=

, 9-x

so

when

it

converges

the

series

converges

to

the

function

f (x)

=

5 9-x

.

17. Does the series

(-1)n(n2 + n3)

n4 + 1

n=1

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

Answer: The series of absolute values

(-1)n(n2 + n3)

n2 + n3

n4 + 1

=

n4 + 1

n=1

n=1

should behave similarly to

n3 n4

,

so

do

a

limit

comparison

to

this

series:

n2 +n3

lim

n

n4 n3

n2 + n3 n4

= lim

n

n4

? n3

n2 + n3

= lim

n

n3

= 1.

n4

Therefore, since

n3 n4

=

values diverges as well.

1 n

diverges,

the

Limit

Comparison

Test

says

that

the

series

of

absolute

However, the given series satisfies the hypotheses of the Alternating Series Test, so it converges. Therefore, since the series converges but the series of absolute values diverges, we conclude that the series converges conditionally.

18. Does the series converge or diverge? Explain your answer.

(2n)! 2n(n!)2

n=1

Answer: Use the Ratio Test:

lim

n

(2(n+1))! 2n+1 ((n+1)!)2

(2n)!

(2n + 2)!

=

lim

n

2n+1((n

+

1)!)2

2n(n!)1 ?

(2n)!

2n (n!)2

(2n + 2)(2n + 1)

(n!)2

= lim

n

2

? ((n + 1)n!)2

(2n + 2)(2n + 1) 1

= lim

n

2

? (n + 1)2 .

Since (2n + 2)(2n + 1) = 4n2 + 6n + 2 and since (n + 1)2 = n2 + 2n + 1, the above limit is equal to

4n2 + 6n + 2 4

lim

n

2(n2

+ 2n + 1)

=

2

=

2.

Since 2 > 1, the Ratio Test implies that the given series diverges.

7

19. Does the series

(-1)n

n ln(n)

n=2

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

Answer: The series of absolute values

(-1)n

1

=

n ln(n)

n ln(n)

n=2

n=2

diverges (see HW #12, Problem 3 for a proof of this). However, the series satisfies the hypotheses of the Alternating Series Test and hence converges, so we see that it converges conditionally.

20. For what values of p does the series

1

(n2 + 1)p

n=0

converge? Explain your answer.

Answer: For large n, the +1 will barely contribute, so do a limit comparison with the series

1 (n2 )p

:

lim

n

1 (n2 +1)p

1 (n2 )p

1

(n2)p

=

lim

n

(n2

+

1)p

?

1

= lim

n

n2 n2 + 1

p

= 1p = 1,

since

limn

n2 n2 +1

=

1.

Therefore, the given series and the series

1 (n2 )p

=

1 n2p

will

either

both

converge

or

both

diverge.

Since

1 n2p

converges

for

2p > 1

and

diverges

otherwise,

we

see

that

the

given

series

converges

when

2p > 1, which is to say when

1 p> .

2

21. What is the interval of convergence of the following power series? Explain your answer.

3n(x - 2)n

n2

.

n=1

Answer: Start by applying the Ratio Test to the series of absolute values:

lim

n

3n+1 (x-2)n+1 (n+1)2

3n (x-2)n n2

3n+1|x - 2|n+1

n2

n2

= lim

n

(n + 1)2

?

3n|x - 2|n

=

lim 3|x - 2| ?

n

(n + 1)2

= 3|x - 2|.

Therefore, the ratio test implies that the given series converges absolutely when 3|x - 2| < 1, meaning

when

|x

-

2|

<

1 3

.

Now,

check

the

endpoints.

When

x-2

=

1 3

,

the

series

becomes

3n

1 3

n2

n

=

1 n2 ,

n=1

n=1

which converges.

8

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