§12.5 –Alternating Series
Mark Woodard
Furman U
Fall 2010
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
1/7
Outline
1
Definitions and theorem
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Definition
Let {an } be a sequence of positive numbers. Infinite series of the form
∞
X
(−1)n+1 an = a1 − a2 + a3 − a4 + · · ·
n=1
∞
X
(−1)n an = −a1 + a2 − a3 + a4 − · · ·
n=1
are called alternating series.
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Definition
Let {an } be a sequence of positive numbers. Infinite series of the form
∞
X
(−1)n+1 an = a1 − a2 + a3 − a4 + · · ·
n=1
∞
X
(−1)n an = −a1 + a2 − a3 + a4 − · · ·
n=1
are called alternating series.
Disclaimer
We will concentrate on alternating series of the form a1 − a2 + a3 − · · · .
All of our results apply to the series −a1 + a2 − a3 + · · · as well.
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
3/7
Definitions and theorem
Theorem
Let {an } be a decreasing sequence of positive terms. If an → 0, then the
alternating series
∞
X
(−1)n+1 an
n=1
converges.
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§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Proof.
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§12.5 –Alternating Series
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Definitions and theorem
Proof.
A simple argument shows that the partial sums satisfy:
· · · s2 ≤ s4 ≤ s6 ≤ · · · s7 ≤ s5 ≤ s3 ≤ s1
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
5/7
Definitions and theorem
Proof.
A simple argument shows that the partial sums satisfy:
· · · s2 ≤ s4 ≤ s6 ≤ · · · s7 ≤ s5 ≤ s3 ≤ s1
In other words, the even numbered terms converge to a limit se and
the odd numbered terms converge to a limit so . We need to show
that these are the same.
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
5/7
Definitions and theorem
Proof.
A simple argument shows that the partial sums satisfy:
· · · s2 ≤ s4 ≤ s6 ≤ · · · s7 ≤ s5 ≤ s3 ≤ s1
In other words, the even numbered terms converge to a limit se and
the odd numbered terms converge to a limit so . We need to show
that these are the same.
But
|so − se | = lim |s2n+1 − s2n | = lim a2n+1 = 0,
n→∞
n→∞
which proves the claim.
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
5/7
Definitions and theorem
Problem
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Problem
∞
X
Show that
(−1)n+1
n=1
Mark Woodard (Furman U)
1
converges.
2n − 1
§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Problem
∞
X
Show that
(−1)n+1
n=1
1
converges.
2n − 1
Leibniz showed that the answer is π/4.
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
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Definitions and theorem
Theorem (Alternating Series Estimation Theorem)
P
n+1 a = a − a + a − a + · · · satisfies
If an alternating series ∞
n
1
2
3
4
n=1 (−1)
0 ≤ an+1 ≤ an and an → 0 as n → ∞, then if s is the sum of the series
and sn is the nth partial sum, then
|Rn | = |s − sn | ≤ an+1 .
Mark Woodard (Furman U)
§12.5 –Alternating Series
Fall 2010
7/7