1.3.5 1.3.6 Ratio TEST Alternating Series TEST PDF

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Download 1.3.5 1.3.6 Ratio TEST Alternating Series TEST PDF

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1.3.5 Ratio Test Theorem: Let ∑ 𝑘=1 𝑢𝑘 be a series with positive terms, and suppose 𝑢𝑘+1 | 𝑢𝑘

𝜌 = lim | 𝑘→∞

a) If 𝜌 < 1, the series converges absolutely, hence converges b) If 𝜌 > 1 or 𝜌 = +∞ , the series diverges c) If 𝜌 = 1, the test is inconclusive

Example Apply Ratio Test to determine if the series converges or diverges. 2𝑘

1. ∑ 𝑘=1 ( 𝑘! )

𝑢𝑘+1 | 𝑢𝑘

𝜌 = lim | 𝑘→∞

2𝑘+1 (𝑘+1)! 2𝑘

= lim | 𝑘→∞

𝑘!

| 𝑘!

2𝑘+1

= lim | (𝑘+1)! . 𝑘→∞

= lim | ( 𝑘→∞

2𝑘

2 𝑘 . 21 𝑘+1) . 𝑘!

.

2

|

𝑘!

2𝑘

|

= lim | 𝑘+1| = 0 < 1 , 𝑘→∞

Series converges absolutely, hence converges 2. ∑  𝑘=1

(2𝑘)! 4𝑘

𝑢𝑘+1 | 𝑘→∞ 𝑢𝑘

𝜌 = lim |

= lim | 𝑘→∞

=

1

4𝑘 41

.(

4𝑘

2𝑘)!

4𝑘

. (2𝑘)!|

|

(2𝑘+2)(2𝑘+1)(2𝑘)!

= lim | 𝑘→∞

4 𝑘+1

(2𝑘+2)!

= lim | 𝑘→∞

|

(2(𝑘+1))!

= lim | 𝑘→∞

(2(𝑘+1))! 4𝑘+1 (2𝑘)! 4𝑘

4𝑘 41

.(

4𝑘

2𝑘 )!

lim |(2𝑘 + 2)(2𝑘 + 1)|

|

4 𝑘→∞

=∞ > 1,

series diverges

3𝑘

3. ∑  𝑘=1 ( ) 𝑘2

𝑢𝑘+1 | 𝑘→∞ 𝑢𝑘

𝜌 = lim |

3𝑘+1 (𝑘+1)2 3𝑘 𝑘2

= lim | 𝑘→∞

|

3𝑘+1

= lim | 𝑘→∞

(𝑘+1)2

𝑘→∞

(𝑘+1)2

= lim |

𝑘2

. 3𝑘 |

3 𝑘 31

𝑘2

. 3𝑘 |

𝑘2

= 3 lim |(𝑘+1)2 | 𝑘→∞

or alternatively,

𝑘2

= 3 lim |

= 3 lim |𝑘 2 +2𝑘+1 | 𝑘→∞

= 3 lim | 𝑘→∞

=3>1

1

2

1

1+ 𝑘+ 2 𝑘

|

𝑘→∞

| 𝑘+1

𝑘→∞

1+𝑘

= 3 lim |

,

1

series diverges

1

4. ∑  𝑘=1 (2𝑘−1)

𝜌 = lim | 𝑘→∞

𝑢𝑘+1

= lim | 𝑘→∞

𝑢𝑘

|

1 2(𝑘+1)−1 1 2𝑘−1

1

|

= lim |2(𝑘+1)−1 . 𝑘→∞

1

= lim |2𝑘+1 . 𝑘→∞

2𝑘−1

= lim |2𝑘+1 | 𝑘→∞

𝑘→∞

5. ∑  𝑛=1

1 𝑘 1 2+ 𝑘

2−

= lim |

2𝑘−1

2𝑘−1 1

1

|

|=1

,

|

test is inconclusive (test fails)

(𝑛+1)! 𝑒 𝑛. 𝑛

𝑢𝑛+1

𝜌 = lim | 𝑛→∞

𝑢𝑛

| = lim | 𝑛→∞

= lim | 𝑛→∞

(𝑛+2)! 𝑒𝑛+1 . (𝑛+1) (𝑛+1)! 𝑒𝑛 . 𝑛

|

(𝑛 + 2)! 𝑒𝑛. 𝑛 | 𝑒 𝑛+1 . (𝑛 + 1) (𝑛 + 1)!

2

𝑘

1

2

| =3

= lim

(𝑛 + 2) (𝑛 + 1)! 𝑒 𝑛 . 𝑛 | (𝑛 + 1)! 𝑛 𝑒 1 . (𝑛 + 1) 𝑛→∞ 𝑒 (𝑛+2) 𝑛 1 = lim | (𝑛+1) |

=

=

= 6. ∑  𝑛=1 (

3𝑛 . 𝑛! 𝑛2

|

𝑒 𝑛→∞

1

𝑒 𝑛→∞

1

lim |

𝑒 𝑛→∞ ∞ 𝑒

𝜌 = lim | 𝑛→∞

𝑢𝑛

|

𝑛+1

𝑛+2 1+

|

1 𝑛

=∞ > 1 ,

) 𝑢𝑛+1

𝑛2 +2𝑛

lim |

| = lim | 𝑛→∞

= lim | 𝑛→∞

= lim | 𝑛→∞

= lim | 𝑛→∞

3𝑛+1 . (𝑛+1)! (𝑛+1)2 3𝑛 . 𝑛! 𝑛2

series diverges

|

3𝑛+1 . (𝑛+1)!

𝑛2

(𝑛+1)2

3𝑛 31 . (𝑛+1)𝑛! (𝑛+1)2

31 . (𝑛+1) 𝑛2 1 (𝑛+1)2

𝑛→∞

𝑛2 (𝑛+1) (𝑛+1)2

𝑛→∞

𝑛+1

𝑛→∞

1+ 𝑛

= 3 lim | = 3 lim |

= 3 lim |

𝑛2

=∞ > 1,

𝑛

1

|

|

3𝑛 .

|

𝑛!

𝑛2

|

3𝑛 . 𝑛!

|

|

series diverges

1.3.6 ALTERNATING SERIES TEST Alternating Series are series whose terms are alternatively positive and negative. For example: 1

1

1

1

1 − 2 + 3 − 4 +. . . +(−1)𝑛+1 −. .. 𝑛

or

1

1

1

1

−1 + 2 − 3 + 4 −. . . +(−1)𝑛 −. .. 𝑛

The general forms of Alternating Series are: 𝑛 ∑∞ 𝑛=1 (−1) 𝑎𝑛

∞

∑

or

(−1)𝑛+1𝑎𝑛

𝑛=1

where ak’s are assumed to be positive in both cases.

Theorem: An alternating series will CONVERGES if: a)

a1 > a2 > a3 > a4 > ….. > an > …..

b)

𝑛→∞

(a’s decreasing)

and

𝑙𝑖𝑚 𝑎𝑛 = 0

Example: Use Alternating Series Test (AST) to show the series converges. 1.

∑

∑

∞

(−1)𝑘+1

𝑘=1

∞

(−1)𝑘+1

𝑘=1

1 1

1

𝑘 𝑘

a) 1 > 2 >

1

1

=1− + − 2 3 1

> 3

1

1

4

1

+ − ⋯ …. 5

> 5 > ⋯ …. 4 1

b) 𝑙𝑖𝑚 𝑎𝑘 = 𝑙𝑖𝑚

𝑘→∞ 𝑘

𝑘→∞

1

=

1

=0



By AST, the series converges.

2.

∑

∑

∞

(−1)𝑘+1

𝑘=1

∞

(−1)𝑘+1

𝑘=1

5

𝑘+3 𝑘(𝑘+1) 𝑘+3 𝑘(𝑘+1)

=

1

4

1(2)

−

5

2(3)

5

6

+ 3(4 ) − ⋯ …

1

= 2 − 6 + − ⋯ … .. 2

a) 2 > 6 > 2 > ⋯ . … ….

𝑘+3 𝑘→∞ 𝑘(𝑘+1)

b) 𝑙𝑖𝑚 𝑎𝑘 = 𝑙𝑖𝑚 𝑘→∞

= 𝑙𝑖𝑚

𝑘+3

𝑘→∞ 𝑘 2 +𝑘

= 𝑙𝑖𝑚

𝑘→∞

1 3 + 𝑘 𝑘2 1 1+𝑘

=

0+0 1+0

By AST, the series converges.

=0...