Download Infinite Series

7-31-2005
Infinite Series
An infinite series is a sum
a0 + a1 + a2 + · · · + an + · · · .
Notation:
∞
X
k=0
ak = a0 + a1 + a2 + · · · + a n + · · · .
For example,
∞
X
k=0
(−1)k = 1 − 1 + 1 − 1 + 1 − · · · + (−1)n + · · · ,
∞
X
1 1
1
1
= 1 + + + ··· + n + ···,
k
2
2 4
2
k=0
∞
X
1 1
1
1
= 1 + + + · · · + + · · ·.
k
2 3
n
k=1
Addition is not defined for an infinite collection of numbers. I have to define what I mean by the
sum of an infinite series like those above. To do this, I’ll look at the sequence of partial sums. For
a0 + a1 + a2 + · · · + an + · · ·, the partial sums are
s0 = a0 ,
s1 = a0 + a1 ,
s2 = a0 + a1 + a2 ,
..
.
sn = a0 + a1 + a2 + · · · + an .
To say that the sum of the series is S means that the sequence of partial sums converges to S:
lim sn = S.
n→∞
The notation for this is
∞
X
ak = S.
k=0
It is often difficult to compute the sum of an infinite series exactly. However, you can often tell that a
series converges without knowing what it converges to. If necessary, a computer can be used to approximate
the sum of a convergent series.
Example. The decimal representation of a real number is a convergent infinite series. For example,
3.14159265 . . .3 +
4
1
5
1
+
+
+
+ ···.
10 100 1000 10000
Repeating decimals represent rational numbers. I’ll show by example how to convert a repeating decimal
to a rational fraction. Consider 0.272727 . . .. Set x = 0.272727 . . .. Then
100x = 27.272727 . . .,
so
99x = 27,
1
and
x=
27
.
99
Example. (Geometric series) A geometric series has the form
a + ar + ar 2 + ar 3 + · · · + ar n + · · · =
∞
X
ar k .
k=0
The picture below shows the partial sums of the geometric series
1+
1 1 1
+ + + ···.
2 4 8
Notice that the partial sums seem to approach 2.
2
1
1+1/2
1+1/2
1+1/2
+1/4 +1/4+1/8
To find a formula for the sum of a geometric series, I’ll start by computing the nth partial sum. By long
division,
1
r n+1
= 1 + r + r2 + · · · + rn +
.
1−r
1−r
(This will make sense provided that r 6= 1.)
Multiply by a, then move the last term on the right to the left:
n
X
k=0
ar k = a + ar + ar 2 + ar 3 + · · · + ar n =
ar n+1
a
−
.
1−r
1−r
For instance,
3 + 3 · 5 + 3 · 52 + · · · + 3 · 5100 =
What about the infinite series
∞
X
k=0
3
3 · 5101
3 101
5 −1 .
−
=
1−5
1−5
4
ar k = a + ar + ar 2 + ar 3 + · · · + ar n + · · ·?
The series converges if the limit of nth partial sum exists. I need to compute
ar n+1
a
a
a
−
−
lim r n+1 .
=
lim
n→∞ 1 − r
1−r
1 − r 1 − r n→∞
By a result on geometric sequences,
lim r n+1 =
n→∞
0
diverges
2
if |r| < 1
.
if |r| ≥ 1
Hence,
∞
X
ar =
k=0
For instance,
k
(
a
1−r
diverges
if |r| < 1
∞
X
1
1 1
1
= 1+ + + ···+ n + ··· =
2k
2 4
2
k=0
∞
X
k=0
.
if |r| ≥ 1
1
1−
1
2
= 2,
3 · 5k = 3 + 3 · 5 + 3 · 52 + · · · + 3 · 5n + · · · diverges.
What about something like
k
5
6
7
∞
X
1
1
1
1
4· −
=4· −
+4· −
+4· −
+ · · ·?
3
3
3
3
k=5
Well, this is
5
1
5
5 5 2
4· −
1
1
1
1
1
3
1
=− .
·1+4· −
· −
· −
+··· =
+4· −
4· −
1
3
3
3
3
3
81
1− −
3
How would you use the formula for the infinite series to find, e.g.,
1000
X
k=100
1
?
2k
Example. (Retirement) $200 is deposited each month and collect 4.8% annual interest, compounded monthly.
How much is in the account after 30 years?
Note that 30 years is 360 months.
4.8% annual interest, compounded monthly, means that each month the amount in the account earns
4.8%
= 0.4% interest. This means that the amount in the account is multiplied by 1.004 each month.
12
The table below tracks each monthly deposit. The first row represents the first $200 deposited, the
second row the second $200 deposited, and so on.
1 month
200
2 months
3 months
...
360 months
1.004 · 200
2
...
1.004 · 200
...
1.004359 · 200
200
1.004 · 200
200
...
200
1.004358 · 200
1.004357 · 200
..
.
1.0041 · 200
200
The total amount in the account is the sum of the numbers in the last column, which is
200 ·
359
X
n=0
1.004n = 200 ·
1 − 1.004360
≈ 160429.
1 − 1.004
3
By comparison, with no interest — e.g. if you put $200 a month under your mattress — you’d only
have $72000 after 30 years.
At the same time, this is a rather sobering conclusion. Many people would find it a challenge to put
away $200 a month toward retirement. This problem shows that doing so and assuming a conservative
interest rate produces a significant total — but hardly enough to retire on!
The following theorem says, roughly, that the sum of two series is the series of their sum, and that
constants may be pulled out of series. These are familiar properties of limits, derivatives, and integrals.
Theorem. Suppose that
∞
X
ak and
∞
X
(ak + bk ) =
∞
X
k=1
ak +
k=1
k=1
(b)
∞
X
c · ak = c ·
∞
X
bk converge, and that c is a constant. Then:
k=1
k=1
(a)
∞
X
∞
X
bk .
k=1
ak .
k=1
Example. Consider the series
converge. Now
n ∞ X
2
1
. This is the sum of two geometric series, both of which
+
n
4
3
n=0
∞
X
1
=
4n
n=0
4
=
1
3
1−
4
1
and
∞ n
X
2
n=0
3
=
1
1−
2
3
= 3.
n ∞ X
1
2
Therefore,
+
converges, and
n
4
3
n=0
n ∞ X
4
13
1
2
= +3=
+
.
n
4
3
3
3
n=0
Example. If each of two series diverge, the sum series may still converge. For example, here are two divergent
series:
∞
∞
X
X
(−1) = −∞.
1 = +∞ and
n=1
n=1
But their sum converges:
∞
X
(1 + (−1)) =
∞
X
0 = 0.
n=1
n=1
Is this happening because “∞ − ∞ = 0”? No! Here are two divergent series:
∞
X
n = +∞
and
∞
X
(−1) = −∞.
n=1
n=1
Now the sum series diverges:
∞
X
n=1
(n + (−1)) =
∞
X
n=1
(n − 1) = 0 + 1 + 2 + · · · = +∞.
4
No conclusion can be drawn in general about the sum of divergent series.
Example. (Telescoping series) Find
∞
X
k=0
2
.
k 2 + 6k + 8
By partial fractions,
2
1
1
=
−
.
k 2 + 6k + 8
k+2 k+4
Then
∞ X
k=0
1
1
−
k+2 k+4
=
All of the fractions except for
1 1
−
2 4
+
1 1
−
3 5
+
1 1
−
4 6
+
1 1
−
5 7
+
1 1
−
6 8
+ ···.
1
1
and cancel. Hence,
2
3
∞
X
k=0
1 1
5
2
= + = .
k 2 + 6k + 8
2 3
6
In many cases, it’s hard to find the exact value of the sum of a convergent series. However, if you know
the series converges, you can approximate the sum of the series as closely as you wish by adding up enough
terms. Thus, in most of what follows, I’ll consider convergence tests, which are methods for determining
whether a series converges or diverges (without necessarily finding a sum).
The first test is fundamental: It says that in a convergent series, the terms must go to 0.
Theorem. (Zero Limit Test) If the series
∞
X
ak converges, then lim ak = 0.
k→∞
k=1
Proof. Suppose the series converges to a sum S. Choose some positive number (junk). I’m going to show
that if k is large enough, then ak must lie within (junk) of 0.
1
To do this, find a number n so that sk and sk−1 are within · junk of S when k ≥ n. (I can do this
2
because the series converges to S, so eventually all the terms get arbitrarily close to S.) Then
|sk − S| + |S − sk−1 | <
1
1
· junk + · junk = junk.
2
2
But by the Triangle Inequality,
|sk − S| + |S − sk−1 | ≥ |sk − S + S − sk−1 | = |sk − sk−1 | = |ak |.
Therefore,
junk > |ak |
for k ≥ n.
Now I didn’t specify the value of (junk) — it can be any positive number. So I’ve just shown that if I
take k to be large enough, I can make ak smaller than any positive number. But this means that the terms
must go to 0:
lim ak = 0.
k→∞
5
You usually use the contrapositive of the theorem, which says:
If
lim ak 6= 0,
k→∞
Example. Apply the Zero Limit Test to
∞
X
k=1
then
∞
X
ak
diverges.
k=1
5k
.
7k + 3
5k
5
= 6= 0,
k→∞ 7k + 3
7
lim
so the series diverges, by the Zero Limit Test.
Example. Apply the Zero Limit Test to
∞
X
cos k.
k=1
lim cos k
k→∞
does not exist,
so the series diverges, by the Zero Limit Test.
Warning: A standard mistake is to use the Zero Limit Test backward. It is not true that if lim ak = 0,
k→∞
then the series
example.
∞
X
k=1
∞
X
1
, is discussed in the next
ak converges. A counterexample, the harmonic series
k
k=1
Example (The harmonic series) The series
∞
X
1 1
1
1
= 1+ + +···+ +···
k
2 3
n
k=1
is called the harmonic series.
1
Draw the graph of y = , 1 ≤ x ≤ n + 1. Divide the interval 1 ≤ x ≤ n + 1 into n + 1 equal subintervals
x
of length 1, and build a rectangle on each subinterval using the left-hand endpoints for the heights.
6
The areas of the rectangle are 1,
1
1
, . . . , . Their sum is
2
n
sn = 1 +
1 1
1
+ +···+ ,
2 3
n
which is the n-th partial sum of the harmonic series.
Since the rectangles lie above the curve,
1 1
1
sn = 1 + + + · · · + >
2 3
n
Take the limit as n → ∞:
Z
n+1
1
1
n+1
dx = [ln x]1 = ln(n + 1).
x
lim sn > lim ln(n + 1) = +∞.
n→∞
n→∞
Therefore, limn→∞ sn = +∞: The harmonic series diverges.
By comparison,
∞
X
1
1
1
1
π2
=
+
+
+
·
·
·
=
.
k2
12
22
32
6
k=1
P∞ 1
I’ll have more to say about series of the form k=1 p below.
k
Finally, go back to the inequality above. I have
1+
Look at
1 1
1
+ + · · · + > ln(n + 1) > ln n.
2 3
n
1
1 1
− ln n.
an = 1 + + + · · · +
2 3
n
It’s possible to show that limn→∞ an converges. The limit is denoted γ, and is called Euler’s constant:
1 1
1
γ = lim 1 + + + · · · + − ln n .
n→∞
2 3
n
γ ≈ 0.5772156649 . . .. So, for example, the sum of the first one million terms of the harmonic series is
1+
1 1
1
+ + · · · + 6 ≈ ln 106 + γ ≈ 13.23830.
2 3
10
Question: Is γ irrational? The answer is not known.
Here are a couple of neat points about the last example.
First, the method I used — that of comparing a series to an integral — can be used on series of the
form
∞
X
1
, p > 0.
kp
k=1
These are called p-series; the harmonic series is the case p = 1. A p-series converges if p > 1 and
diverges if 0 < p ≤ 1. For example,
∞
X
1
converges,
k3
k=1
∞
X
1
√ diverges,
k
k=1
7
∞
X
k −3/4 diverges.
k=1
If p > 1, the sum of the p-series is denoted ζ(p). Thus,
ζ(3) =
∞
X
1
.
k3
k=1
It isn’t too difficult to find closed form expressions for ζ(2n), where n is an integer. For instance,
ζ(2) =
∞
X
π2
1
=
.
2
k
6
k=1
However, the odd sums ζ(2n + 1) are somewhat mysterious. It was only in 1978 that R. Apéry showed
that ζ(3) is irrational. No one knows what its exact value is, and no one knows if (for instance) ζ(5) is
irrational.
Second, the method of comparing a series to an integral works more generally.
∞
X
ak is a series in which the terms are positive. Let f(x) be the function you get by
Theorem. Suppose
k=1
replacing k by x in ak . Suppose that:
(a) f is continuous for x ≥ 1.
(b) f decreases for x ≥ 1.
Then:
If
If
Z
∞
converges, so does
f(x) dx
1
Z
∞
f(x) dx diverges, so does
1
∞
X
k=1
∞
X
ak .
ak .
k=1
This is called the Integral Test. Here’s why it works. Divide the interval [1, n + 1] up into rectangles
of width 1, using the left-hand endpoints to get the heights. The function f decreases, so the picture looks
like this:
y=f(x)
1
2
The sum of the rectangle areas is the n
curve:
3
th
n n+1
partial sum, and it is clearly bigger than the area under the
sn = f(1) + f(2) + · · · + f(n) >
Therefore, if lim
Z
n→∞ 1
Z
n+1
f(x) dx.
1
n+1
f(x) dx diverges, so does lim sn , but lim sn is the sum of the series.
n→∞
8
n→∞
Next, divide the interval [1, n] up into rectangles of width 1, but use the right-hand endpoints to get
the heights. The picture looks like this:
y=f(x)
1
2
3
n
The sum of the rectangle areas is clearly smaller than the area under the curve:
Z n
f(2) + f(3) + · · · + f(n) <
f(x) dx.
1
If I add f(1) to both sides, the left side becomes the n-th partial sum:
Z n
sn = f(1) + f(2) + f(3) + · · · + f(n) < f(1) +
f(x) dx.
1
If lim
Z
n→∞ 1
n
f(x) dx converges, so does lim sn , but lim sn is the sum of the series.
n→∞
n→∞
Example. Think of using the Integral Test when the general term of the series looks like something that you
can integrate.
∞
X
Consider the series
ke−k . xe−x is something you can integrate, so it’s natural to try the Integral
k=1
Test.
For k ≥ 1, ke−k > 0. The series has positive terms.
Let f(x) = xe−x. f is continuous for x ≥ 1. Since
f ′ (x) = (1 − x)e−x < 0 for x > 1,
f decreases for x ≥ 1. The hypotheses of the Integral Test are satisfied. (It’s important to check the
hypothesis before applying the test!)
Compute the improper integral:
Z ∞
Z c
c
−x
xe dx = lim
xe−x dx = lim −xe−x − e−x 1 =
c→∞
1
c→∞
1
lim −ce−c − e−c + e−1 + e−1 = 2e−1 .
c→∞
Here’s the parts table:
Z
d
dx
+
x
−
1
+
0
dx
e−x
ց
ց
→
9
−e−x
e−x
Since the integral converges, the series converges by the Integral Test.
Warning: The value of the integral in the Integral Test is not equal to the sum of the series.
The proof of the Integral Test yields the formula
f(1) +
Z
n
f(x) dx > sn >
1
Z
n+1
f(x) dx.
1
You can use this to estimate the partial sums of a series to which the Integral Test applies.
Example. Estimate the sum of the first 1000 terms of
∞
X
1
√
.
3
k
k=1
This is a divergent p-series with p =
f(1) +
Z
1000
1
Now
f(1) +
Z
1
1000
1
x1/3
1
, so the Integral Test applies. I have
3
1
x1/3
dx > s1000 >
dx ≈ 149.5 while
Z
1001
1
Z
1001
1
Thus, 148.59998 < s1000 < 149.5.
c 2005 by Bruce Ikenaga
10
1
x1/3
dx.
1
dx ≈ 148.59998.
x1/3