Download Series: Infinite Sums

Series: Infinite Sums
Series are a way to make sense of certain types of infinitely long sums. We will
need to be able to do this if we are to attain our goal of approximating transcendental functions by using ‘infinite degree’ polynomials. But before we try to add
together an infinite number of polynomials, we first explore what it means to add
an infinite number of numbers.
Here’s the issue: We know how to add two numbers: a1 + a2 . Using associativity
(and parentheses) we can add three numbers
a1 + ( a2 + a3 )
four numbers
a1 + ( a2 + ( a3 + a4 ))
or even n numbers
a1 + ( a2 + ( a3 + ( a4 + (· · · + ( an
1
+ an ) . . . )))).
But where would we start (or end) when trying to add an infinite number of
terms? And does the sum add up to a finite number or not? Since all we know
how to do is add a finite number of terms, we will have to use finite addition and
limits to make sense of the process.
Introduction to Series
OK, enough of this finite stuff. What we want to do is add up the terms of an
•
infinite sequence { an }•
n=1 . More precisely, given a sequence { an }n=1 , we can form
the infinite sum
a1 + a2 + a3 + · · · + a n + · · · =
•
 ak
k =1
which is called an infinite series or more simply just a series.
Can we do this? Here are several examples.
(a)
•
 k = 1 + 2 + 3 + 4 + · · · = •. The sum is clearly not finite; the series
k =1
diverges.
•
(b)
1
1 1 1
= 1 + + + + · · · . Do these terms add up to a finite sum?
k
2 3 4
k =1
(c)
1
1 1 1
= 1 + + + + · · · . Do these terms add up to a finite sum?
n
2
2 4 8
n =0
(d)
Â
•
Â
•
1
=1
( 2) n
n =0
Â
1 1
+
2 4
1
+ · · · . Do these terms add up to a finite sum?
8
math 131
infinite series, part vii: absolute and conditional convergence
DEFINITION 13.1. To find the sum of an infinite series •
k=1 ak we form the sequence of
partial sums that are often denoted by Sn .
S1 = a 1
S2 = a 1 + a 2
S3 = a 1 + a 2 + a 3
..
.
Sn = a1 + a2 + a3 + · · · + a n =
n
 ak
(Sn is called the nth partial sum of the series)
k =1
If the sequence of partial sums {Sn }•
n=1 has a limit a limit L (converges), we say that the
series converges to L and we write:
•
n
k =1
k =1
a = lim Sn = L
 ak = nlim
!• Â k
n!•
or just
•
 ak = L.
k =1
Otherwise the series diverges.
EXAMPLE 13.1. Here’s a simple example. Find the sum of the series
•
1
1 1
1
1
= + ++ +
+··· ,
k
2
4
8
16
2
k =1
Â
if it exists.
SOLUTION. We first determine each partial sum and then rewrite it in a more conve-
nient form.
1
1
=1
2
2
1 1
3
1
S2 = + = = 1
2 4
4
4
1 1 1
7
S2 = + + = = 1
2 4 8
8
..
.
S1 =
1
8
1 1 1
1
1
+ + +··· n = 1
2 4 8
2
2n
n
o•
1
So the sequence of partial sums is {Sn }•
and
n =1 = 1
2n
Sn =
n =1
1
=1
2n
lim Sn = lim 1
n!•
n!•
0 = 1,
where we have used Theorem 13.2 to evaluate the limit. In other words,
•
1
= 1.
k
k =1 2
Â
Pretty cool!
EXAMPLE 13.2. Here’s a another fun example. Find the sum of the series
it exists.
•
1
if
2+k
k
k =1
Â
SOLUTION. Using partial fractions (check this)
•
•
1
 k2 + k = Â
k =1
k =1
✓
1
k
1
k+1
◆
=
✓
1
1
2
◆
+
✓
1
2
1
3
◆
+
✓
1
3
1
4
◆
+
✓
1
4
1
5
◆
+···
2
3
Notice that most of the terms cancel out. The sum collapses and we see that
✓
◆ ✓
◆ ✓
◆ ✓
◆
✓
◆
1
1 1
1 1
1 1
1
1
1
Sn = 1
+
+
+
+···+
=1
.
2
2 3
3 4
4 5
n
n+1
n+1
Such a sum is called a telescoping sum. We are left with only the first and last terms
in the partial sum. This time
1
=1
n+1
lim Sn = lim 1
n!•
n!•
In other words,
0 = 1.
•
1
= 1.
2+k
k
k =1
Â
◆
1
1
 k k + 2 if it
k =1
exists. This time there will be a few more terms that do not cancel. See if you can figure it
out.
YOU TRY IT 13.1. Try this telescoping sum. Find the sum of the series
•
✓
EXAMPLE 13.3 (Partial Fractions). Here’s another example that uses partial fractions.
Find the sum of the series
•
4
if it exists.
2 + 3k + 2
k
k =0
Â
SOLUTION. Since the degree of the numerator is smaller than the degree of the de-
nominator a nd since the denominator factors into linear factors, we can write
4
4
A
B
Ak + 2A + Bk + B
=
=
+
=
.
(k + 1)(k + 2)
k+1
k+2
(k + 1)(k + 2)
k2 + 3k + 2
Solving we get:
k’s :
0 = A + B.
(13.1)
and
constants :
4 = 2A + B.
(13.2)
4 = A.
(13.3)
Subtracting (13.1) from (13.2) gives
Putting A = 4 in (13.1) makes B =
k2
4. So we see that
4
4
=
k
+
1
+ 3k + 2
4
.
k+2
(Check that this is correct!) This means that
•
• ✓
4
4
=
 k2 + 3k + 2  k + 1
k =0
k =0
4
k+2
◆
which is another telescoping series. This time
✓
◆ ✓
◆ ✓
◆
✓
4 4
4 4
4 4
4
Sn =
+
+
+···+
1 2
2 3
3 4
n+1
So
lim Sn = lim 4
In other words,
n!•
n!•
4
=4
n+2
4
n+2
◆
=4
4
.
n+2
0 = 4.
•
4
= 4.
2 + 3k + 2
k
k =0
Â
Wow!
EXAMPLE 13.4 (Telescoping). Here’s a more complicated example that uses partial
fractions. Find the sum of the series
•
 ln
k =1
( k + 1)
if it exists.
k
math 131
infinite series, part vii: absolute and conditional convergence
SOLUTION. We can use a log property to rewrite the partial sum as
Sn =
n
 ln
k =1
✓
k+1
k
◆
n
=
 ln(k + 1)
ln k
k =1
= (ln 2
= ln n
ln 1) + (ln 3
ln 2) + (ln 4
ln 3) + · · · + [ln(n + 1)
ln 1.
ln n]
Therefore
ln 1 = • (diverges)
lim Sn = lim ln n
and the series
n!•
n!•
•
 ln
k =1
diverges.
✓
k+1
k
◆
EXAMPLE 13.5 (Partial Fractions). Here’s a more complicated example that uses partial
fractions. Find the sum of the series
•
Â
k =0
k2
8
if it exists.
+ 4k + 3
SOLUTION. Since the degree of the numerator is smaller than the degree of the de-
nominator a nd since the denominator factors into linear factors, we can write
8
4
A
B
Ak + 3A + Bk + B
=
=
+
=
.
(k + 1)(k + 3)
k+1
k+3
(k + 1)(k + 3)
k2 + 4k + 3
Solving we get:
k’s :
0 = A + B.
(13.4)
and
constants :
8 = 3A + B.
(13.5)
8 = 2A.
(13.6)
Subtracting (13.4) from (13.5) gives
Putting A = 4 in (13.4) makes B =
4. So we see that
8
4
=
k+1
k2 + 4k + 3
This means that
4
.
k+3
• ✓
8
4
=
Â
2 + 4k + 3
k
+
1
k
k =0
k =0
which is another telescoping series.
⇣
⌘ ⇣
⌘ ⇣
⌘ ⇣
Sn = 41 43 + 42 44 + 43 45 + 44
⇣
⌘ ⇣
4
4
· · · + n 4 1 n+
1 + n
•
Â
Sn = 4 + 2
4
n +2
4
n +3 .
So
lim Sn = lim 6
In other words,
n!•
n!•
4
n+2
4
k+3
4
5
◆
⌘
+···
⌘ ⇣
4
4
n +2 + n +1
4
n +3
⌘
4
= 6.
n+3
•
8
= 6.
2 + 4k + 3
k
k =0
Â
YOU TRY IT 13.2 (Partial fractions). Here are two others that are similar to the last example
in that they use partial fractions. See if you can solve them. Find the sums of these series if
they exist.
•
1
2 + 7k + 12
k
k =0
•
1
(b) Â 2
k
+
4k
+3
k =0
(a)
Â
Answers: (a) 14 ; (b) 32 .
4
5
Geometric Series
Geometric series are among the simpler with which to work. We will see that we
can determine which ones converge and what their limits are fairly easily.
DEFINITION 13.2. A geometric series is a series that has the form
•
 arn , where a is a real
n =0
constant and r is a real number.
YOU TRY IT 13.3. Here are a few examples. Identify a and r in each.
•
 6·4
(a)
n
n =0
•
1
(b) Â n
2
n =0
•
n
 2·3
(c)
(d)
n =0
 5·
n =2
Determining the sum of a geometric series
✓
2
3
◆n
✓ ◆2n
1
(e) Â 2 ·
3
n =1
•
 arn is relatively simple. We begin
n =0
by comparing the nth partial sum Sn with rSn . We find:
Sn = a + ar + ar2 + ar3 + · · · + ar n
rSn =
2
3
n
ar + ar + ar + · · · + ar + ar
(13.7)
n +1
(13.8)
So subtracting (13.8) from (13.7) we obtain
Sn
ar n
rSn = a
1
or
(1
r n +1 ).
r ) Sn = a (1
So
r n +1 )
.
1 r
We know from the Key Limit Theorem 13.2 that
8
>
if |r | < 1
>0
<
n
lim r = 1
if r = 1
n!•
>
>
:
diverges otherwise.
Sn =
a (1
(13.9)
(13.10)
Thus, putting (13.9) and (13.10) together we find
8
< a
if |r | < 1
a (1 r n +1 )
lim Sn = lim
= 1 r
n!•
n!•
:diverges otherwise.
1 r
So we have proved
THEOREM 13.1 (Geometric Series Test). If |r | < 1, then the geometric series
and
•
 arn =
n =0
If |r |
1, then the geometric series
a
1
r
•
 arn converges
n =0
.
•
 arn diverges.
n =0
EXAMPLE 13.6. Here are some examples that get progressively more complex.
(a)
Find the sum of the series
✓ ◆n
2
 5 if it exists.
n =0
•
Answers: (a) 6 and 4; (b) 1 and 1/2; (c)
2 and 1/3; (d) a = 20/9 and r = 2/3;
(e) a = 2/9 and r = 1/9.
math 131
infinite series, part vii: absolute and conditional convergence
SOLUTION. In this example a = 1 and r = 25 and |r | < 1. So by Theorem 13.1 the
series converges to
1
1
= 53 .
2
5
(b) Find the sum of the series
•
Â4
n =0
✓ ◆n
6
if it exists.
7
SOLUTION. In this example a = 4 and r = 67 and |r | < 1. So by Theorem 13.1 the
series converges to
(c)
1
4
=
6
7
28
1
= 28.
✓ ◆n
3
Find the sum of the series  2
if it exists.
2
n =0
•
SOLUTION. In this example a = 2 and r = 32 . Since |r | > 1, by Theorem 13.1 the
series diverges.
(d) Find the sum of the series
•
Â
5
n =0
✓
1
2
◆ n +2
if it exists.
SOLUTION. Before we can apply the Geometric Series Test, we have to adjust the
power. Notice that we can rewrite the series using the nth power using
•
Â5
n =0
Now a =
5
4
1 (
1
2)
=
5
4
5
4
3
2
✓
1
2
◆ n +2
1
2
and r =
=
5
6.
•
=
Â5
n =0
✓
1
2
◆2 ✓
◆n
1
2
•
5
= Â
4
n =0
✓
1
2
◆n
.
and |r | < 1. So by Theorem 13.1 the series converges to
SOLUTION. Alternative Method. Another way that we can approach this prob-
lem is to write out the first few terms of the series and identify a and r.
•
Â5
n =0
Now a =
5
4
✓
1
2
◆ n +2
5
4
|{z}
=
5
5
+
8
16
|{z} |{z}
a
ar
ar2
ar3
1
2
and the ratio of a term to the previous one is r =
5
4
by Theorem 13.1 the series converges to
(e)
5
+··· .
32
|{z}
Find the sum of the series
Â
2
n =2
1 (
=
1
2)
5
4
3
2
=
5
6.
and |r | < 1. So
I find this easier!
✓ ◆n
1
if it exists.
3
SOLUTION. We use the Alternative Method. Write out the first few terms of the
series and identify a and r.
✓ ◆n
1
2
2
2
2
 2 3 = 9 + 27 + 81 + 243 + · · · .
n =2
|{z} |{z} |{z} |{z}
ar
a
Now a =
2
9
ar2
ar3
and the ratio of a term to the previous one is r =
Theorem 13.1 the series converges to
2
9
1 ( 13 )
=
2
9
2
3
= 13 .
1
3
and |r | < 1. So by
Some series start at n = 1 or n = 2 or some other index n 6= 0. For instance,
consider the general series
•
 an . We can write it as a series starting at n = 0 if we
n =2
are careful to also subtract off the first two terms:
!
•
•
n =2
n =0
 an =  an
We can apply this to geometric series.
( a0 + a1 ).
6
7
EXAMPLE 13.7. Here are a two more examples.
(a)
Find the sum of the series
✓ ◆n
2
if it exists.
3
n =3
•
Â
SOLUTION. First rewrite the series adding back and then subtracting the first few
‘missing’ terms.
✓ ◆n !
2
 3
n =0
✓ ◆n
2
 3 =
n =3
•
✓
•
By Theorem 13.1 the series converges to
1
1
⇣
2
3
1+
1+
2
3
2 4
+
3 9
4
9
+
⌘
◆
.
19
9
=3
= 89 .
Alternative Method. Write out the first few terms of the series and identify a and
r.
• ✓ ◆n
2
8
16
32
64
 3 = 27 + 81 + 243 + 729 + · · · .
n =3
|{z} |{z} |{z} |{z}
a
Now a =
8
27
ar
ar2
ar3
and the ratio of a term to the previous one is r =
8
27
2
3
8
9.
by Theorem 13.1 the series converges to 1 ( 2 ) =
3
✓
◆
•
1 n
(b) Find the sum of the series  4
if it exists.
3
n =2
and |r | < 1. So
SOLUTION. First rewrite the series; be careful of the signs.
•
Â
n =2
4
✓
1
3
◆n
•
Â
=
4
n =0
✓
4
1 (
By Theorem 13.1 the series converges to
1
3
◆n !
⇣
1
3)
✓
4
4
3
4
4
3
⌘
◆
=3
.
8
3
= 13 .
Alternative Method. Write out the first few terms of the series and identify a and
r.
✓
◆
•
1 n
4
4
4
4
+
+··· .
Â4 3 = 9
27
81
243
n =2
|{z} |{z} |{z} |{z}
ar
a
Now a =
4
9
ar2
ar3
and the ratio of a term to the previous one is r =
Theorem 13.1 the series converges to
4
9
1 (
1
3)
=
1
3.
1
3
and |r | < 1. So by
(Optional) Application: Repeating Decimals
You may (or may not) remember from your high school math days that every
repeating decimal can be expressed as a rational number, that is, as a fraction
using integers. There are some familiar ones such as
0.3333 · · · = 0.3 =
Similarly we have
1
.
3
2
0.6666 · · · = 0.6 =
3
and
0.1111 · · · = 0.1 =
1
.
9
Note: We use a horizontal bar to
indicate which part of the decimal
repeats.
math 131
infinite series, part vii: absolute and conditional convergence
But what about something like 0.12 = 0.121212 . . . ? We can write any such expression as a geometric series. In this case
0.12 = 0.121212 . . . = 0.12 + 0.0012 + 0.000012 + · · ·
12
12
12
12
+
+
+
+···
2
6
4
(10)
(10)
(10)8
(10)
✓
◆
12
1
1
1
=
1
+
+
+
+
·
·
·
(10)2
(10)2 (10)4 (10)6
!
✓
◆2 ✓
◆3
12
1
1
1
=
1+
+
+
+···
(10)2
(10)2
(10)2
(10)2
!
✓
◆
✓
◆
12
1
1 2
1 3
=
1+
+
+
+···
100
100
100
100
✓
◆
•
12
1 n
= Â
100 100
n =0
=
=
=
=
12
100
1
1
100
12
100
99
100
12
.
99
So 0.12 = 0.121212 . . . is rational since it can be written as the fraction
12
99 .
EXAMPLE 13.8. Here are a few more repeating decimals to try.
(a)
Express 0.9 = 0.9999 . . . as a rational number. What is interesting about the
answer?
(b) Express 0.123 as a rational number.
(c)
Express 0.abcd as a rational number, where a, b, c, and d are nonnegative integers.
(d) Express 0.1234 as a rational number.
Answers: (b)
123
999 ;
(c)
abcd
9999 ;
(d)
1222
9900 .
8
Basic Properties of Series
The Algebra of Series
There are several simple properties that apply to series that converge. We have
already been using the first of these results.
THEOREM 14.1. Suppose that we have two convergent series:
•
 an
= A and
n =0
Then
(a)
•
 bn
= B.
n =0
•
If c is a constant, then
 can = cA.
n =0
(b)
•
 an ± bn = A ± B.
n =0
EXAMPLE 14.1. Determine the sum of the series
•
Â
3
n =0
✓ ◆n
✓
◆
1
1 n
+3
if it exists.
2
2
SOLUTION. Using the Geometric Series Test (Theorem 13.1) and Theorem 14.1 we
have
•
Â
n =0
3
✓ ◆n
✓
◆
✓ ◆n
✓
◆
•
•
1
1
1 n
1 n
+3
= Â3
+Â3
2
2
2
2
n =0
n =0
3
3
=
+
1 12
1 ( 12 )
= 6 + 2 = 8.
The Divergence Test
The next theorem shows that for a series to converge, the terms of the series must
get small as n gets large.
THEOREM 14.2 (The nth term Test). If
•
an = 0.
 an converges to A, then nlim
!•
n =0
Proof. We make use of a comment we made earlier. Let Sn be the nth partial sum
of the series. We know that lim Sn = lim Sn 1 = A because both sequences are
n!•
n!•
really just the same set of numbers. But
Sn = a1 + a2 + · · · + a n
and
Sn
1
= a1 + a2 + · · · + a n
So
Sn
Sn
1
= an .
1
+ an
1.
math 131
infinite series, part vii: absolute and conditional convergence
10
Therefore
lim an = lim (Sn
n!•
Sn
n!•
1)
= lim Sn
lim Sn
n!•
1
n!•
A = 0,
=A
which is what we wanted to prove.
The theorem is seldom used in the form as stated above. Rather, if the nth term
of a series does not converge to 0, then series cannot convergence. This is a more
useful way of stating of Theorem 14.2 and in this form it is a test for divergence.
THEOREM 14.3 (The nth term test for divergence). If lim an 6= 0, then
n!•
•
 an diverges.
n =0
Warning! The nth term test for divergence never allows us to conclude that a
series converges, only that it does not converge. If lim an = 0, we can’t conclude
n!•
anything. The series may or may not converge. The test simply fails to provide us
with any useful information in such a case.
EXAMPLE 14.2. Determine whether the series
•
n2 + 1
converges.
+n+1
Â
3n2
n =0
SOLUTION. Notice that
1 + n12
n2 + 1
=
lim
n!• 3n2 + n + 1
n!• 3 + 1 +
n
lim an = lim
n!•
1
n
=
By the nth term test for divergence (Theorem 14.3), the series
verges.
EXAMPLE 14.3. Determine whether the series
•
Â
n =1
✓
1+
k
n
◆n
1
6= 0.
3
•
n2 + 1
di2
n=0 3n + n + 1
Â
converges.
SOLUTION. This time using using one of our key limits (see Theorem 13.2)
lim an = lim
n!•
n!•
✓
1+
k
n
◆n
= ek 6= 0.
By the nth term test for divergence (Theorem 14.3), the series
•
Â
n =1
EXAMPLE 14.4. Determine whether the series
•
Â
p
n
✓
1+
k
n
◆n
diverges.
n converges.
n =1
SOLUTION. Using another of our key limits (see Theorem 13.2)
lim an = lim
n!•
n!•
p
n
n = 1 6= 0.
By the nth term test for divergence (Theorem 14.3), the series
•
Â
p
n
n diverges.
n =1
EXAMPLE 14.5. Determine whether the series
•
2n
2 4 8
= + + + · · · converges.
2
1 4 9
n
k =1
Â
SOLUTION. Notice that both numerator and denominator both tend to infinity. So
converting to x and using l’Hôpital’s rule (twice!)
lim
n!•
2n
2x
(ln 2)2x
(ln 2)(ln 2)2x
=
lim
=
lim
=
lim
= •.
x !• x2
x !•
x !•
2x
2
n2
By the nth term test for divergence (Theorem 14.3), the series diverges.
We will need to use the derivative
d
formula dx
( a x ) = a x ln a, which is valid
when a > 0.
11
EXAMPLE 14.6. Determine whether the series
SOLUTION. Notice that both
lim
n!•
and
•
•
1
1
and  2 converge.
n
n
k =1
k =1
Â
1
= 0.
n
1
= 0.
n2
In this situation the nth term test for divergence (Theorem 14.3) fails to provide us
with any information. The series may or may not converge. In fact, we will soon see
that one of these series converges while the other diverges.
lim
n!•
EXAMPLE 14.7. Determine whether the series
•
 cos n1
converges.
k =1
SOLUTION. Notice that
lim cos
n!•
1
n
= cos 0 = 1 6= 0.
By the nth term test for divergence (Theorem 14.3) the series diverges.
YOU TRY IT 14.4. Here are six series. Which of them can you say diverge by the nth term
test for divergence? For which series is this test not helpful. Explain.
•
3n + 1
2n + 5
n =1
✓
◆
•
4 n
(d) Â 1 +
n
n =1
(a)
Â
•
n
2+1
2n
n =1
• p
(e) Â n n
(b)
Â
n =1
(c)
(f )
•
 (1.1)n
n =1
•
n!
n
n
n =1
Â