SolutionP - Memorial University of Newfoundland

Assignment 2 answer key

1. Determine which of the following series is divergent and explain why.

Solution:

We will use the statement that if limn¡ú¡Þ an 6= 0 then series

P¡Þ

a

is

divergent.

n=1 n

¡Þ

X

n+1

1

n+1

; lim

= 6= 0. The series diverges.

2n ? 1 n¡ú¡Þ 2n + 1

2

a)

n=1

b)

¡Þ

X

3n

n=1

n3

; By L¡¯Hospital¡¯s Rule

3n

(ln 3)3n

(ln 3)2 3n

(ln 3)3 3n

lim 3 = lim

=

lim

=

lim

=¡Þ

n¡ú¡Þ n

n¡ú¡Þ

n¡ú¡Þ

n¡ú¡Þ

3n2

6n

6

The series diverges.

c)

¡Þ

X

2n

, geometric series with r=2. The series diverges.

100

n=1

d)

¡Þ

X

(1 +

n=1

k n

) ,

n



For all k the limit lim

n¡ú¡Þ

¡Þ

X

k

1+

n

n

= ek 6= 0. The series diverges.

n ?2n

n ?2n

n + 3 2n

)

. The limit lim (

)

lim (

) = e6 6= 0.

n¡ú¡Þ n + 3

n¡ú¡Þ

n+3

n

n=1

The series diverges.

e)

(

2. For each of the following series, find the sum of the convergent series.

Solution

All of the following series are geometric related. Here we use

P¡Þ

n = a .

ar

n=0

1?r

¡Þ

X





¡Þ 

¡Þ 

X

7 n X 9 n

(a)

[(0.7) + (0.9) ] =

+

?2

10

10

n=1

n=0

n=0

1

1

10

34

=

7 +

9 ? 2 = 3 + 10 ? 2 = 3

1 ? 10

1 ? 10

 n

¡Þ

X

1

?1

4

8

4

(b) 4 ? 2 + 1 ? + ... =

4

=

= 3 =

1

2

2

3

1 ? (? 2 )

2

n=0

(c)

¡Þ

X

n=0

n

(?2)(

n

?2 n

?2

6

) =

=? .

3

5

1 ? (? 32 )

1

3. Find the values of x for which the series converges, and find the sum

of the series for those values of x.



¡Þ 

¡Þ

¡Þ

X

(4x ? 1)n X 4x ? 1 n X n

4x ? 1

=

=

(a)

r ; where r =

n

5

5

5

n=0

n=0

n=0

4x ? 1

? The series converges iff |r| < 1 ? |

| 2.

The integral can be evaluated by parts

R ¡Þ 2 ?x

P¡Þ 2 ?n

5

?x 2

¡Þ

n=1 n e

1 x e dx = ?e (x +2x+2)|1 = e . Thus the series

converges.

¡Þ

X

Similarly, for k = 3. We have

n3 e?n ; Then f (x) = x3 e?x is

n=1

positive (for x > 0)and continuous. To check that the function is

decreasing we find its derivative f 0 (x) = e?x (?x3 + 3x) and show

that f 0 (x) < 0 for all x > 3.

The integral can be evaluated by parts

R ¡Þ 3 ?x

16

?x 3

2

¡Þ

1 x e dx = ?e (x + 3x + 6x + 6)|1 = e . Thus the series

P

¡Þ

3 ?n converges.

n=1 n e

For any integer k ¡Ý 1 we have: f (x) = xk e?x is positive (for

x > 0)and continuous. To check that the function is decreasing

we find its derivative f 0 (x) = e?x (?xk + kxk?1 ) and show that

f 0 (x) < 0 for all x > k.

UsingZ integration by parts and mathematical induction we show

¡Þ

that

xk e?x dx =

1

Z ¡Þ

h

i¡Þ

1 k k(k ? 1)

k!

+¡¤¡¤¡¤+ .

= ?xk e?x + k

xk?1 e?x dx = + +

e

e

e

e

1

1

Thus the series converges for any k ¡Ý 1.

¡Þ

X

1

(b)

.

1/3

n

n=1

1

Function f (x) = 1/3 is positive, continuous and decreasing for

x

x

0.





Z>

¡Þ

1

3 2 A

x3

dx = lim

= ¡Þ. Thus the series diverges.

A¡ú¡Þ 2

x1/3

1

1

(c)

(d)

¡Þ

X

1

Convergent p-series with p = ¦Ð >1.

n¦Ð

n=1

¡Þ

X

ln n

. Function f (x) =

n3

n=2

f 0 (x)

ln x

x3

1?3 ln x

x4

is continous and positive for x ¡Ý 2.

< 0 for x > e1/3 . Thus the function is decreasing.





Z ¡Þ

Z t

ln x

1 ln x

1 t

f (x)dx = lim

dx

=

lim

?

?

t¡ú¡Þ 2 x3

t¡ú¡Þ

2 x2

4x2 2

2





? ln t

1

ln 2

1

1 + 2 ln 2

1

ln t

= lim

? 2+

+

=

? lim 2 ? lim 2 =

2

t¡ú¡Þ

t¡ú¡Þ 4t

t¡ú¡Þ 2t

2t

4t

8

16

16

1 + 2 ln 2

1 + ln 4

?0?0=

? the series is convergent.

16

16

=

3

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