4-3: Alternating Series, and the Alternating Series Theorem
Prakash Balachandran
Department of Mathematics
Duke University
February 5, 2010
• Due to some short term-amnesia over break, I’ve decided that you’ll be allowed a one page cheat sheet for
this exam only (on an 8 12 × 11 piece of paper). Use this opportunity to review the essential material from
31L.
• Let’s begin with a prototype example:
∞
X
(−1)k+1
k=1
k
.
Does it converge?
• Clearly none of the tests we’ve developed so far can help us (I dare you to try them). So how can we
determine if this series converges?
P
k+1
• Idea: let’s look at Sn = nk=1 (−1)k for n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
• Obviously, if we continue, we find the “hopping” behavior continues infinitely often. But, as we go higher
in n, the magnitude of each additional “hop” is getting smaller and smaller by n1 , so that we eventually settle
down to a value as n → ∞ (if you’re curious, the above series converges to ln 2 which we’ll prove later in
the semester).
• Definition: A series is called
Pan alternating series if the terms alternate in sign. That is, an alternating
series is a series of the form (−1)k+1 ak where ak > 0 for all k.
• The series above is thus an example of an alternating series, and is called the alternating harmonic series.
• The idea of hopping back and forth to a limit is basically the proof of:
Theorem 1 (The Alternating Series Theorem)
The alternating series
∞
X
(−1)n−1 an = a1 − a2 + a3 − a4 + · · · + (−1)n−1 an + · · ·
n=1
converges if 0 < an+1 < an for all n and limn→0 an = 0.
• As usual, once we have a test for convergence, we ask: what’s the error in approximating the convergent
sum by an nth partial sum? The answer is in the
P
Theorem 2 (Error Estimate for Alternating Series) Let Sn = ni=1 (−1)i−1 ai be the nth partial sum of an
alternating series and let S = limn→∞ Sn . Suppose that 0 < an+1 < an for all n and limn→∞ an = 0.
Then, |S − Sn | < an+1 .
Proof of Theorem 2:
1
– S − Sn =
P∞
i−1
ai .
i=n+1 (−1)
P
i−1
= −an+1 + ∞
ai > −an+1
i=n+2 (−1)
P∞
P
i−1
– When n is even: 0 < S − Sn = i=n+1 (−1)i−1 ai = an+1 + ∞
ai < an+1
i=n+2 (−1)
– When n is odd: 0 > S − Sn =
P∞
i−1
ai
i=n+1 (−1)
– Both of these say that |S − Sn | < an+1 . • Examples:
P∞
(−1)n−1
np
converges for any p > 0. Contrast this with the p-series.
P
(−1)k−1
by S9 ? What does this tell us about the interval
– What’s the error in approximating ln 2 = ∞
k=1
k
in which S lies?
– The series
n=1
∗ By the error estimate for alternating series: |S − S9 | < a10 =
1
.
10
∗ From my calculator: S9 = 0.7456 so that
−
1
1
1
1
< S − S9 <
⇔ S9 −
< S < S9 +
⇔ 0.6456 < S < 0.8456
10
10
10
10
and so S lies somewhere in the interval (0.6456, 0.8456) (ln 2 ≈ 0.6931).
P
n n2
– Does ∞
n=1 (−1) 1+n2 converge?
∗ This diverges by the nth term test.
P
– Does ∞
n=1 sin(ln n) converge?
∗ Ditto.
P
– Does ∞
n=1
(−1)n
1
ln(1+ n
)
converge?
∗ Ditto.
• Definition: A series
P
ak is called conditionally convergent if
P
|ak | diverges, but
P
ak converges.
• (∗) A remarkable theorem of Cauchy shows that if you have a conditionally convergent series, then I can
add all the terms up in a different order to obtain any number I want in the limit. Obviously this contrasts
to what happens in the case of summing a finite number of numbers.
– The general proof is too hard for this class, but let’s get a flavor for it by using the alternating harmonic
P
(−1)n−1
series: ∞
.
n=1
n
– Let’s isolate the positive and negative terms:
1 1 1
1, , , , . . . = {a1 , a3 , a5 , a7 , . . .}
3 5 7
1 1 1 1
− , − , − , − , . . . = {a2 , a4 , a6 , a8 , . . .} .
2 4 6 8
– Now, here’s the idea: I give you a number x. For the sake of simplicity, suppose x > 0. Start
adding the positive numbers a1 , a3 , . . . until I just get above x and stop. Say I add M1 terms, and let
b1 = a1 , b2 = a3 , . . . , bM1 = a2M1 +1 .
P
P∞
1
Note: I can always do this since ∞
n=1 a2n−1 =
n=1 2n−1 diverges.
2
– Now start adding the negative terms a2 , a4 , . . . until I just pass below x and stop. Suppose I add M2
terms, and let bM1 +1 = a2 , bM1 +2 = a4 , . . . bM1 +M2 = a2M2 .
P
P∞ 1
Note: Again, I can always do this since ∞
n=1 a2n = −
n=1 2n diverges.
– Now start adding the positive terms a2M1 +3 , a2M1 +5 , . . . until I just get above x and stop. Rinse and
repeat.
– The idea is to continue getting just above x and just below x. The result is a new sequence of terms
bk from the original sequence ak that are in a different order. By the construction of bk , its nth partial
sums will continue to hop just above x and just below x, so that since the terms I’m adding are getting
closer and closer to 0, I hit x in the limit.
Awesome!!!!!!!!!!!
3