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10.6 The Comparison, Ratio, and Root Tests

THEOREM (The Comparison Test): Letk=1

ak andk=1

bk be series with nonnegative terms and

suppose that a1 b1, a2 b2, a3 b3, . . . , ak bk, . . .(a) If the bigger series

k=1

bk converges, then the smaller seriesk=1

ak also converges.

(b) If the smaller series k=1

ak diverges, then the bigger series k=1

bk also diverges.

EXAMPLE: Use the comparison test to determine whether the following series converge or diverge:

(a)k=2

13

k 1(b)

k=1

1

k2 +k+ 1 (c)

k=2

1

k2 1SOLUTION:

(a) Since 13

k 1>

13

kand

k=2

13

kdiverges, it follows that

k=2

13

k 1also diverges.

(b) Since 1

k2 +k+ 1 0, then the series both converge or both diverge.

EXAMPLE: Use the limit comparison test to determine whether the following series converge ordiverge:

(a)

k=1

1

k+ 1 (b)

k=2

1

2k2 k 1 (c)k=1

4k2

k+ 5

k5 +k4 + 2k 2SOLUTION:

(a) Put an= 1

k+ 1, bn=

1k

. Then = limk+

ak

bk= lim

k+

k

k+ 1= lim

k+

1

1 + 1k

= 1. Since

= 1 andk=1

1k

diverges, it follows thatk=1

1k+ 1

also diverges.

(b) Putan= 1

2k2 k 1 , bn= 1

2k2.Then= lim

k+

ak

bk= lim

k+

2k2

2k2 k 1= limk+1

1 12k 1

2k2

=

1. Since = 1 and

k=2

1

2k2 =

1

2

k=2

1

k2 converges, it follows that

k=2

1

2k2 k 1 also converges.

(c) Put an = 4k2 k+ 5k5 +k4 + 2k 2 , bn =

4

k3. Then = lim

k+

ak

bk= lim

k+

k3(4k2 k+ 5)4(k5 +k4 + 2k 2) =

limk+

4k5 k4 + 5k34k5 + 4k4 + 8k 8= limk+

1 14k

+ 54k2

1 + 1k

+ 2k4 2

k5

= 1.Since = 1 andk=1

4

k3 = 4

k=1

1

k3 converges,

it follows thatk=1

4k2 k+ 5k5 +k4 + 2k 2 also converges.

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THEOREM (The Ratio Test): Let

uk be a series with positive terms and suppose that

= limk+

uk+1

uk.

(a) If 1 or = +, the series diverges.(c) If= 1, the series may converge or diverge, so that another test must be tried.

EXAMPLE: Use the ratio test to determine whether the following series converge or diverge:

(a)k=1

k+ 1

k! (b)

k=1

k

3k+1 (c)

k=1

(2k)k+2

(k+ 1)! (d)

k=1

(2k)!

3k (e)

k=1

1

3k+ 4

SOLUTION:

(a) We have = limk+

uk+1

uk= lim

k+

k+ 2

(k+ 1)! k!

k+ 1 = lim

k+

k+ 2

(k+ 1)(k+ 1) = 0 1, therefore the series di-

verges.

(d) We have = limk+

uk+1

uk= lim

k+

(2k+ 2)!

3k+1 3

k

(2k)!= lim

k+

(2k+ 1)(2k+ 2)

3 = +,therefore

the series diverges.

(e) We have = limk+

uk+1

uk= lim

k+

3k+ 4

3k+ 7= 1, therefore the series may converge or diverge, so

that another test must be tried. For example, by the integral test,

+

1

dx

3x+ 4 = lim

+

1

dx

3x+ 4 =

lim+

1

3ln(3x+ 4)

1

= +, therefore the series diverges.

THEOREM (The Root Test): Let

uk be a series with positive terms and suppose that

= limk+

k

uk

(a) If 1 or = +, the series diverges.(c) If= 1, the series may converge or diverge, so that another test must be tried.

EXAMPLE: Use the root test to determine whether the following series converge or diverge:(a)

k=2

4k 52k+ 1

k(b)

k=1

1

(ln(k+ 1))k

SOLUTION:

(a) We have = limk+

k

uk = limk+

4k 52k+ 1

= 2> 1, therefore the series diverges.

(b) We have = limk+

k

uk = limk+

1

ln(k+ 1)= 0< 1, therefore the series converges.