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QUIZ 4 FALL 2009

Q1. Investigate the convergence or divergence of4 3

3

1.

2k k k

Solution. I. Let 4 31

2ka

k k and 4

1kb

k. Then

4 3 4

4 4 3

1/( 2 )lim lim lim 1 0

1/ 2

k

k k kk

a k k k

b k k k

.

Since4

3

1

k k

is a convergentp-series (p = 4) it follows by limit comparison Test

that the given series converges.

II. By comparison: 4 3 4 4 423

1 1 3

2k k k k k (for all 3k ). Now since

4 43 3

3 13

k kk k

and 43

1

k k

is a convergentp-series (p = 4), it follows by

comparison that the given series is convergent.

Q2. Investigate the convergence or divergence of2

1

3.

6k

k

k k

Solution. Let2

3( ) .

6

xx

x x

Then2

3( ) 0

6

xx

x x

(for all 1x ).

Moreover,2 2

2 2 2 2

( 6 )(1) ( 3)(2 6) 6 18'( ) 0

( 6 ) ( 6 )

x x x x x xx

x x x x

,

for all 1x . Hencef(x) is decreasing, and so by the integral test we have2

2 2

1 1 7

3 1 ( 6 ) 1

6 2 6 2

x d x x dtdx

x x x x t

(

2 6t x x )

7

lim lim ln lim[ln ln 7]7

R

R R RRdt t R

t .

Therefore the given series diverges.

The decreasing part could also be shown by letting2

3,

6k

ka

k k

and so for all

1,k

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2

1

2

( 4) /[( 1) 6( 1)] ( 4)( 6)1

( 3) /( 6 ) ( 1)( 3)( 7)

k

k

a k k k k k k

a k k k k k k

1 ,k ka a

showing that ka is decreasing.

Q3. Given the series1

k

k

a

, if lim 0k

ka

, what conclusion can you make about

convergence or divergence of the series?

Solution. None, because the series may converge or diverge. Consider thep-series

withp = 1 andp = 2. Both satisfy lim 0kk

a

yet the first series diverges while the

second series converges.

QUIZ 4 FALL 2010

Q1. Is the series convergent or divergent?

Solution. Let . Then for all

Now

However, is a convergent 2-series. Thus, by comparison is convergent,and hence is convergent.

Q2. Is the series convergent or divergent?

Solution. Let for all Then for all we see that

is continuous.

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Hence we can apply the Integral Test:

Therefore the given series is convergent since the integral converges.

Q3. Answer TRUE or FALSE(a)If the series is convergent then is convergent also. TRUE

(b)If the series is convergent then lim 0kk

a

. TRUE

(c)In the Ratio Test for series ifL = 1 then the series is divergent. FALSE

Section 8.1. # 23. Determine the convergence or divergence of

.Soultion 1. First express the general term of the sequence as follows:

and then consider the associated function

Now apply LHopitals Rule to get the limit.

Solution 2. First observe that for all n > 0, . [You can prove this byinduction.] Thus by comparison we have for all n > 0,

as Since 2.99/3 < 1.