28 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 Σn=1

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

n=1

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

Alternating Series and Conditional Convergence

An alternating series is a series with alternating positive and negative terms. Alternating series usally

are given as Σn=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 as

Σn=1

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

Alternating Series and Conditional Convergence

An alternating series is a series with alternating positive and negative terms. Alternating series usally

are given as Σn=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 and lim an = 0, then the

alternating series converges. n∞

From here on, we will write an alternating series as

Σn=1

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

Σn=1

∞an

Alternating Series and Conditional Convergence

An alternating series is a series with alternating positive and negative terms. Alternating series usally

are given as Σn=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 and lim an = 0, then the

alternating series 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 as

Σn=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.

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11

S21/2

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11

S21/2

S31/3

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11

S21/2

S31/3

S41/4

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

0 1

S11

S21/2

S31/3

S41/4

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

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

∞

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.

Let Sn = be the partial sum. 12

13 ...1 + 1

4 – – 1

n±

We may visualize Sn geometrically as following:

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)

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an converges.

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an converges. However, to be precise, we define

two notions of convergence for alternating series:

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an

If also converges, we say the alternating series

converges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define

two notions of convergence for alternating series:

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an

If also converges, we say the alternating series

converges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define

If also diverges, we say the alternating series

converges conditionally.

Σn=1

|an|∞

two notions of convergence for alternating series:

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an

If also converges, we say the alternating series

converges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define

If also diverges, we say the alternating series

converges conditionally.

Σn=1

|an|

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

n=1

∞

∞

two notions of convergence for alternating series:

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an

If also converges, we say the alternating series

converges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define

If also diverges, we say the alternating series

converges conditionally.

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

∞

∞

two notions of convergence for alternating series:

Alternating Series and Conditional Convergence

Σn=1

∞Thus given an alternating sequence an 0,

an

If also converges, we say the alternating series

converges absolutely.

Σn=1

|an|∞

converges. However, to be precise, we define

If also diverges, we say the alternating series

converges conditionally.

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

∞

∞

two notions of convergence for alternating series:

1rn,Σ

n=1

∞ 1npΣ

n=1

∞

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

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

n=1

∞

conditionally (because diverges).1npΣ

∞

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

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

n=1

∞

conditionally (because diverges).1npΣ

n=1

∞

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

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

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

n=1

∞

conditionally (because diverges).1npΣ

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).

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

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

n=1

∞

conditionally (because diverges).1npΣ

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 …

Alternating Series and Conditional Convergence

The alternating harmonic series converges1n(-1)nΣ

n=1

∞

conditionally (because diverges).1nΣ

n=1

∞

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

n=1

∞

conditionally (because diverges).1npΣ

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.

Alternating Series and Conditional Convergence

(-1)n

Example:

b. an = 3n n2 + n – 1.

Alternating Series and Conditional Convergence

(-1)n Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

Alternating Series and Conditional Convergence

(-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.

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominating terms are 3n n2 .

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

Compare this to an.

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

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

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

Compare this to an.

n∞Lim 3n

n2 + n – 1 3n n2

= 1

3n n2

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

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

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

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.

Since an 0, converges.Σn=1

∞an

Example:

b. an = 3n n2 + n – 1.

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

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

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.

Since an 0, converges.Σn=1

∞an

∞

Example:

b. an = 3n n2 + n – 1.

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

Alternating Series and Conditional Convergence

(-1)n

We need to identify what type of convergence it is.

The dominate terms are 3n n2 .

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

Since an 0, converges.Σn=1

∞

an

Σn=1

3n n2 + n – 1

(-1)n

∞

∞converges conditionally.

Alternating Series and Conditional Convergence

Summary: Given an alternating sequence {an}

Alternating Series and Conditional Convergence

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

Alternating Series and Conditional Convergence

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|.

Alternating Series and Conditional Convergence

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

If it converges, the alternating series converges absolutely.

Σn=1

∞|an|.

Alternating Series and Conditional Convergence

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

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

Σn=1

∞|an|.