1.2 INFINITE Series PDF

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1.2

INFINITE SERIES

1.2.1 Definition and Notation An infinite series is an expression of the form: a1 + a2 +a3 + a4 + ……+ an + ….. In summation notation, 𝑎1 + 𝑎2 + 𝑎 3 + 𝑎4 + ⋯ = ∑  𝑛=1 𝑎𝑛 = 𝑆𝑛 i.

A series ∑  𝑛=1 𝑎𝑛 converges if lim 𝑆𝑛

= lim (∑ 𝑛=1 𝑎𝑛) = 𝑘, where 𝑘 is a real

A series ∑ 𝑛=1 𝑎𝑛 diverges if lim 𝑆𝑛

= lim (∑ 𝑛=1 𝑎𝑛) =   . A divergent

𝑛→∞

𝑛→∞

number. The value of 𝑘 is called the limit or sum of the series. ii.

𝑛→∞

series has no limit or sum.

𝑛→∞

Example: Determine if the series converges or diverges. converges. 1. ∑ 𝑛=1 𝑎𝑛 =

Find the sum if the series

𝑛2 +1

𝑛2 +2

𝑛2 +1

lim 𝑆𝑛 = lim ( 𝑛2 +2) 𝑛→∞

𝑛→∞

1+1 ⁄ 2 𝑛 ) 𝑛→∞ 1+2 ⁄𝑛2

= lim (

=1

Hence, the series converges to limit 1

2. ∑  𝑛=1 (1 + 𝑛) ∑ 𝑛=1 (1 + 𝑛) = (1 + 1) + (1 + 2) + (1 + 3) + (1 + 4) + ⋯ … . . Thus, 𝑆1 = 2 𝑆1 = 2 𝑆2 = 5 𝑆3 = 9 𝑆4 = 14 Since 𝑆→  as 𝑛 →  , then lim 𝑆𝑛 =  Hence, the series diverges. 1

1

3. ∑  𝑛=1 ( 2𝑛−1 − 2𝑛+1) = 1 − 2

2

2𝑛+1

lim 𝑆𝑛 = lim (1 − 2𝑛+1) = 1

𝑛→∞

𝑛→∞

𝑛→∞

Hence, series converges to limit 1.

4. 1 + (−1) + 1 + (−1) + ⋯ … + (−1)𝑛−1 + ⋯ .. From the series, we notice that 1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑 𝑆𝑛 = { 0 , 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 Hence, lim 𝑆𝑛 does not exist. Series diverges. 𝑛→∞

1.2.2 Telescoping Sum of Series Telescoping series is a series of the form where its sum can be found when nearly every term cancel with either a succeeding or preceding term. In many cases, the sum 𝑆𝑛 is not given, so we need to find it first. Example 1. Determine whether the series converge or diverge. If converges, find the sum. 1

∑ 𝑛=1 (

𝑛(𝑛+1)

(note that 𝑢𝑛 is given, but not 𝑆𝑛 )

)

To find 𝑆𝑛 , we rewrite the 𝑢𝑛 , 𝑢𝑛 = =

1

𝑛(𝑛+1) 𝐴

𝑛

+

𝐵

𝑛+1

Solve, and get 𝐴 = 1, 𝐵 = −1 Hence, 𝑢𝑛 = 1

∑ 𝑛=1 (

𝑛(𝑛+1) 1

1

𝑛

1

− 𝑛+1

) 1

= ∑ 𝑛=1 ( 𝑛 − 𝑛+1) 1

1

1

1

1

1

1

1

= ( − ) + ( − ) + ( − ) + ( − ) + ⋯ .. 1 2 3 4 2 3 4 5 1

1

1

1

1

1

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

1

= 1− 𝑛+1 = 𝑆𝑛 1

Now, lim 𝑆𝑛 = lim (1 − 𝑛+1 ) = 1 − 𝑛→∞

𝑛→∞

Hence, series converges to sum 1

1



=1

2. Use Telescoping series to show that: 3

3 ) 5𝑘+2

∑ 𝑘=1 ( 𝑘 − 5

3

∑ 𝑘=1 (5𝑘 − 3

3

5𝑘+2

3

=

18

25

) 3

3

3

3

3

3

3

3

= ( 5 − 53 ) + (52 − 4 ) + ( 53 − 5 ) + (54 − 6 ) + ( 55 − 7 ) + ⋯ … … .. 5 5 5 5 3

… … … … + ( 5𝑛−4 − 3

=5+

3

52

3

3

3 ) 5𝑘+2

3

3

3

5𝑛

3

) + ( 5𝑛−1 −

3

5𝑛+1

3

3

) + ( 5𝑛 − 5𝑛+2 )

= 𝑆𝑘

5𝑛+2

= lim 𝑆𝑘 𝑘→∞

3

3

3

= lim ( 5 + 52 − 5𝑛+1 − =

3

) + ( 5𝑛−3 − 5𝑛−1 ) + ( 5𝑛−2 −

3

− 5𝑛+1 −

∑ 𝑘=1 ( 𝑘 − 5

3

5𝑛−2

𝑘→∞ 18

3

5𝑛+2

)

25

1.2.3 Geometric Series A geometric series (GS) is a series of the form: ∑ 𝑛=1 𝑎𝑟 𝑛 = 𝑎 + 𝑎𝑟 + 𝑎𝑟 2 + 𝑎𝑟 3 + 𝑎𝑟 4 + ⋯ . . +𝑎𝑟 𝑛−1 + ⋯ … ..

(𝑎  0)

The series will: i.

converges if |r| < 1. It has the sum of 𝑆 =

ii.

diverges if |r| ≥ 1

𝑎 . 1−𝑟

Example 1. Determine whether the series converges or diverges. If converges find its sum. a) 0.6 + 0.06 + 0.006+…… This is a GS with a=0.6 and r=0.1. Since |r| < 1, the series converges and has the sum of 𝑆=

𝑎

1−𝑟

=

0.6

1−0.1

= 0.666

2

2

2

2 3𝑛−1

b) 2 + + 2 + 3 + ⋯ … . . + 3 3 3

+ ⋯ .. 1

This is a GS with a=2 and 𝑟 = 3. Since |r| < 1, the series converges and has the sum of 𝑆=

𝑎

1−𝑟

2

=

1−1⁄3

=3

𝑘 c) ∑  𝑘=0 3

The series is 30 + 31 + 32 + 33 + 34 + ⋯ … which is a GS with 𝑎 = 1 and 𝑟 = 3 Since |r|≥ 1, the series diverges

1 𝑘

d) ∑  𝑘=1 (− 2)

1

1

1

The series is − 2 + 4 − + ⋯ … 8 1

1

which is a GS with 𝑎 = − 2 and 𝑟 = − 2. Since |r| < 1, the series converges and has the sum of 𝑆=

𝑎

=

1−𝑟

2𝑘 1−𝑘 e) ∑  𝑘=1 3 5 𝑘 = ∑ 𝑘=1 (9

9

51

5𝑘

= ∑ 𝑘=1 5 (5 )

1

−2 1 1−(− ) 2

1

= −3

)

𝑘

which is a GS with 𝑟 =

9 5

Since |r|≥ 1, the series diverges

∑ 𝑘=1 (

2. Find the sum of the series:  Let ∑ 𝑘=1 𝑢𝑘 = ∑ 𝑘=1 (

=

3

4

3

4𝑘

)

3

+ 42 +

which is a GS with 𝑎 =

3

43 3

4

3

4𝑘

−

2

5𝑘−1

)

+⋯… 1

and 𝑟 = . 4

Since |r| < 1, the series converges and has the sum of 𝑆=

𝑎

1−𝑟

=

3 4

1 4

1−( )

=1

 Let ∑ 𝑘=1 𝑣𝑘 = ∑ 𝑘=1 (

2

5𝑘−1

2

) 2

= 2 + 5 + 52 +

2

53

+⋯… 1

which is a GS with 𝑎 = 2 and 𝑟 = 5. Since |r| < 1, the series converges and has the sum of 𝑎

𝑆 = 1−𝑟 =

2

1

1−(5)

=

5

2

Hence, ∑ 𝑘=1 (

3

4𝑘

2

− 5𝑘−1 ) = ∑𝑘=1(𝑢𝑘 − 𝑣𝑘 ) = ∑ 𝑘=1(𝑢𝑘 ) − ∑ 𝑘=1(𝑣𝑘 ) 5

3

=1−2=−2

EX. 9.2, No. 9-15...