x28 alternating series and conditional convergence

Alternating Series and Conditional Convergence

Alternating Series and Conditional Convergence An alternating series is a series with alternating positive and negative terms.

Alternating Series and Conditional Convergence An alternating series is a series with alternating positive and negative terms. Alternating series usally are given as Σ

n=1

∞(-1)nan = -a1 + a2 – a3 + a4 …. or as Σ

n=1

∞(-1)n+1an = a1 – a2 + a3 – a4 … where an > 0


n=1

∞(-1)nan = -a1 + a2 – a3 + a4 …. or as Σ

n=1

∞(-1)n+1an = a1 – a2 + a3 – a4 … where an > 0

From here on, we will write an alternating series asΣn=1

∞an where the sequence {an} has alternating signs.


n=1

∞(-1)nan = -a1 + a2 – a3 + a4 …. or as Σ

n=1

∞(-1)n+1an = a1 – a2 + a3 – a4 … where an > 0

Theorem (Alternating Series): Suppose {an} is a alternating sequence where lim an = 0 and that |an+1| ≥ | an | for all large n, then converges.

n∞

We will use the alternating harmonic series

to demonstrate the argument for the theorem.

12

13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1

4 – –

From here on, we will write an alternating series asΣn=1

∞an where the sequence {an} has alternating signs.

Σn=1

∞an

Alternating Series and Conditional Convergence 12

13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1

4 – –The convergence of

may be demonstrated on the real line.


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1


may be demonstrated on the real line. Let Sn = be the partial sum. 1

213 ...1 + 1

4 – – 1

n±


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±


0 1

S11

S21/2


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±


0 1

S11

S21/2

S31/3


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±


0 1

S11

S21/2

S31/3

S41/4


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±


0 1

S11

S21/2

S31/3

S41/4

Since 0, the red dot must approach a limit which is

1n

Σn=1

(-1)n+1 . 1n

∞


13 ...1 +Σ

n=1(-1)n+1 = 1

n∞ 1



213 ...1 + 1

4 – – 1

n±


0 1

S11

S21/2

S31/3

S41/4

Since 0, the red dot must approach a limit which is

1n

Σn=1

(-1)n+1 . 1n

∞(Later we'll see its Ln(2) 0.59)


Σn=1

∞Thus given an alternating sequence an 0,

an converges.


Σn=1


an converges. However, to be precise, we define

two notions of convergence for alternating series:


Σn=1


an

If also converges, we say the alternating seriesconverges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define



Σn=1


an


Σn=1

|an|∞


If diverges, we say the alternating seriesconverges conditionally.

Σn=1

|an|∞



Σn=1


an


Σn=1

|an|∞



Σn=1

|an|

The alternating geometric series , | r | < 1 1rn(-1)nΣ

n=1

∞

∞



Σn=1


an


Σn=1

|an|∞



Σn=1

|an|

The alternating geometric series , | r | < 1

and the alternating p-series where p > 1

converges absolutely

1np(-1)n

1rn(-1)nΣ

n=1

∞

Σn=1

∞

∞



Σn=1


an


Σn=1

|an|∞



Σn=1

|an|

The alternating geometric series , | r | < 1

and the alternating p-series where p > 1

converges absolutely (because converges).

1np(-1)n

1rn(-1)nΣ

n=1

∞

Σn=1

∞

∞


1rn,Σ

n=1

∞ 1npΣ

n=1

∞

Alternating Series and Conditional Convergence The alternating harmonic series converges1

n(-1)nΣn=1

∞

conditionally (because diverges).1nΣ

n=1

∞


n(-1)nΣn=1

∞


n=1

∞

The alternating p-series , p < 1 converges1np(-1)nΣ

n=1

∞

conditionally (because diverges).1npΣ

∞


n(-1)nΣn=1

∞


n=1

∞


n=1

∞


n=1

∞

Hence when we answer questions about the convergence of alternaing series, we always identify if it converges absolutely or condiditonally


n(-1)nΣn=1

∞


n=1

∞


n=1

∞


n=1

∞

Hence when we answer questions about the convergence of alternaing series, we always identify if it converges absolutely or condiditonally(or if it diverges, i.e an doesn't go to 0).


n(-1)nΣn=1

∞


n=1

∞


n=1

∞


n=1

∞

Hence when we answer questions about the convergence of alternaing series, we always identify if it converges absolutely or condiditonally(or if it diverges, i.e an doesn't go to 0). Example: Discuss the convergence of the following theorem. a. (-1)nΣ

n=1

∞= -1 + 1 – 1 + 1 …


n(-1)nΣn=1

∞


n=1

∞


n=1

∞


n=1

∞

Hence when we answer questions about the convergence of alternaing series, we always identify if it converges absolutely or condiditonally(or if it diverges, i.e an doesn't go to 0). Example: Discuss the convergence of the following theorem. a. (-1)nΣ

n=1

∞= -1 + 1 – 1 + 1 … diverges since an does not

converges to 0.

Example:b. an = 3n

n2 + n – 1.


(-1)n

Example:b. an = 3n

n2 + n – 1.


(-1)n Since an 0, converges.Σn=1

∞an

Example:b. an = 3n

n2 + n – 1.


(-1)n

We need to identify what type of convergence it is.

Since an 0, converges.Σn=1

∞an

Example:b. an = 3n

n2 + n – 1.


(-1)n


The dominating terms are 3n n2 .


∞an

Example:b. an = 3n

n2 + n – 1.


(-1)n


The dominate terms are 3n n2 .

Compare this to an.


∞an

Example:b. an = 3n

n2 + n – 1.


(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2


∞an

Example:b. an = 3n

n2 + n – 1.


(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1


∞an

Example:b. an = 3n

n2 + n – 1.

So {an} and are almost-multiple of each other.


(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

3n n2


∞an

Example:b. an = 3n

n2 + n – 1.



(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

3n n2

But Σn=1

∞ 3n n2 = Σ

n=1

∞ 3 n diverges.


∞an

Example:b. an = 3n

n2 + n – 1.



(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

3n n2

But Σn=1

∞ 3n n2 = Σ

n=1

∞ 3 n diverges.

Therefore Σn=1

3n n2 + n – 1

diverges.


∞an

∞

Example:b. an = 3n

n2 + n – 1.



(-1)n



Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

3n n2

But Σn=1

∞ 3n n2 = Σ

n=1

∞ 3 n diverges.

Therefore Σn=1

3n n2 + n – 1

diverges.

So


∞an

Σn=1

3n n2 + n – 1 (-1)n

∞

∞converges conditionally.


Summary: Given an alternating sequence {an}


Summary: Given an alternating sequence {an}1. check if an 0, if no, the series diverges


Summary: Given an alternating sequence {an}1. check if an 0, if no, the series diverges2. if yes, check what type of convergence by investigating the convergence of

Σn=1

∞|an|.



If it converges, the alternating series converges absolutely.

Σn=1

∞|an|.



If it converges, the alternating series converges absolutely. If it diverges, the alternating series converges conditionally.

Σn=1

∞|an|.

x28 alternating series and conditional convergence

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