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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
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
From here on, we will write an alternating series asΣn=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 Σ
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.
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. 1
213 ...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. 1
213 ...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. 1
213 ...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. 1
213 ...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. 1
213 ...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. 1
213 ...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. 1
213 ...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. 1
213 ...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 seriesconverges 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 seriesconverges absolutely.
Σn=1
|an|∞
converges. However, to be precise, we define
If diverges, we say the alternating seriesconverges 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 seriesconverges absolutely.
Σn=1
|an|∞
converges. However, to be precise, we define
If diverges, we say the alternating seriesconverges 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 seriesconverges absolutely.
Σn=1
|an|∞
converges. However, to be precise, we define
If diverges, we say the alternating seriesconverges 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 seriesconverges absolutely.
Σn=1
|an|∞
converges. However, to be precise, we define
If diverges, we say the alternating seriesconverges 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 converges1
n(-1)nΣn=1
∞
conditionally (because diverges).1nΣ
n=1
∞
Alternating Series and Conditional Convergence The alternating harmonic series converges1
n(-1)nΣn=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 converges1
n(-1)nΣn=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 converges1
n(-1)nΣn=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 converges1
n(-1)nΣn=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 converges1
n(-1)nΣn=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|.