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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 transcen-
dental 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: a 1 + a 2. Using associativity
(and parentheses) we can add three numbers
a 1 + ( a 2 + a 3 )
four numbers
a 1 + ( a 2 + ( a 3 + a 4 ))
or even n numbers
a 1 + ( a 2 + ( a 3 + ( a 4 + (· · · + ( a n 1 + a n )... )))).
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 { a n } • n = 1. More precisely, given a sequence { a n } • n = 1 , we can form
the infinite sum
a 1 + a 2 + a 3 + · · · + a n + · · · =
Â
k = 1
a k
which is called an infinite series or more simply just a series.
Can we do this? Here are several examples.
(a)
Â
k = 1
k = 1 + 2 + 3 + 4 + · · · = •. The sum is clearly not finite; the series
diverges.
(b)
Â
k = 1
k
+ · · ·. Do these terms add up to a finite sum?
(c)
Â
n = 0
2 n^
+ · · ·. Do these terms add up to a finite sum?
(d)
Â
n = 0
( 2 ) n^
+ · · ·. Do these terms add up to a finite sum?
math 131 infinite series, part vii: absolute and conditional convergence 2
DEFINITION 13. 1. To find the sum of an infinite series (^) Â • k = 1 a (^) k we form the sequence of partial sums that are often denoted by S (^) n.
S 1 = a (^1) S 2 = a 1 + a (^2) S 3 = a 1 + a 2 + a (^3) .. .
S (^) n = a 1 + a 2 + a 3 + · · · + a (^) n =
n  k = 1
a (^) k ( S (^) n is called the n th partial sum of the series )
If the sequence of partial sums (^) { S (^) n } • n = 1 has a limit a limit L (converges), we say that the series converges to L and we write:
 k = 1
a (^) k = (^) nlim! •
n  k = 1
a (^) k = (^) nlim! • S (^) n = L
or just
 k = 1
a (^) k = L.
Otherwise the series diverges.
EXAMPLE 13. 1. Here’s a simple example. Find the sum of the series
 k = 1
2 k^
if it exists.
SOLUTION. We first determine each partial sum and then rewrite it in a more conve- nient form.
S 1 =
S 2 =
S 2 =
S (^) n =
2 n^
2 n
So the sequence of partial sums is { S (^) n } • n = 1 =
n 1 (^21) n
o (^) • n = 1 and
nlim! • S^ n^ =^ nlim! • 1 ^
2 n^
where we have used Theorem 13. 2 to evaluate the limit. In other words,
 k = 1
2 k^
Pretty cool!
EXAMPLE 13. 2. Here’s a another fun example. Find the sum of the series
 k = 1
k 2 + k if it exists.
SOLUTION. Using partial fractions (check this)
 k = 1
k 2 + k
 k = 1
k
k + 1
math 131 infinite series, part vii: absolute and conditional convergence 4
SOLUTION. We can use a log property to rewrite the partial sum as
S (^) n =
n  k = 1
ln
k + 1 k
n  k = 1
ln ( k + 1 ) ln k
= ( ln 2 ln 1 ) + ( ln 3 ln 2 ) + ( ln 4 ln 3 ) + · · · + [ ln ( n + 1 ) ln n ] = ln n ln 1.
Therefore nlim! • S^ n^ =^ nlim! • ln^ n^ ^ ln 1^ =^ •^ (diverges) and the series (^) • Â k = 1
ln
k + 1 k
diverges.
EXAMPLE 13. 5 (Partial Fractions). Here’s a more complicated example that uses partial fractions. Find the sum of the series
 k = 0
k 2 + 4 k + 3 if it exists.
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 k 2 + 4 k + 3
( k + 1 )( k + 3 )
A
k + 1
B
k + 3
Ak + 3 A + Bk + B ( k + 1 )( k + 3 )
Solving we get: k’s : 0 = A + B. ( 13. 4 ) and constants : 8 = 3 A + B. ( 13. 5 ) Subtracting ( 13. 4 ) from ( 13. 5 ) gives 8 = 2 A. ( 13. 6 )
Putting A = 4 in ( 13. 4 ) makes B = 4. So we see that 8 k 2 + 4 k + 3
k + 1
k + 3
This means that (^) • Â k = 0
k 2 + 4 k + 3
 k = 0
k + 1
k + 3
which is another telescoping series.
S (^) n =
4 1 ^ 4 3
4 2 ^ 4 4
4 3 ^ 4 5
4 4 ^ 4 5
4 n 1 ^ 4 n + 1
4 n ^ 4 n + 2
4 n + 1 ^ 4 n + 3
S (^) n = 4 + 2 (^) n +^4 2 (^) n +^4 .
So lim n! • S (^) n = lim n! •
n + 2
n + 3
In other words,
 k = 0
k 2 + 4 k + 3
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.
(a)
 k = 0
k 2 + 7 k + 12
(b)
 k = 0
k 2 + 4 k + 3 Answers: (a) 14 ; (b) 32.
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
 n = 0
ar n^ , where a is a real
constant and r is a real number.
YOU TRY IT 13. 3. Here are a few examples. Identify a and r in each.
(a)
 n = 0
6 · 4 n^ (b)
 n = 0
2 n^ (c)
 n = 0
2 · 3 ^ n^ (d) (^) Â n = 2
◆ (^) n (e) (^) Â n = 1
◆ (^2) n 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.
Determining the sum of a geometric series
 n = 0
ar n^ is relatively simple. We begin
by comparing the nth partial sum S n with rS n. We find:
S n = a + ar + ar 2 + ar 3 + · · · + ar n^ ( 13. 7 )
rS n = ar + ar 2 + ar 3 + · · · + ar n^ + ar n^ +^1 ( 13. 8 )
So subtracting ( 13. 8 ) from ( 13. 7 ) we obtain
S n rS n = a ar n^ ^1
or
( 1 r ) S n = a ( 1 r n^ +^1 ).
So
S n =
a ( 1 r n^ +^1 )
1 r
We know from the Key Limit Theorem 13. 2 that
lim
n! •
r n^ =
0 if | r | < 1
1 if r = 1
diverges otherwise.
Thus, putting ( 13. 9 ) and ( 13. 10 ) together we find
lim
n! •
S n = lim
n! •
a ( 1 r n^ +^1 )
1 r
a
1 r if^ |^ r^ |^ <^1
diverges otherwise.
So we have proved
THEOREM 13. 1 (Geometric Series Test). If | r | < 1, then the geometric series
 n = 0
ar n^ converges
and (^) •
 n = 0
ar n^ = a 1 r
If | r | 1, then the geometric series
 n = 0
ar n^ diverges.
EXAMPLE 13. 6. Here are some examples that get progressively more complex.
(a) Find the sum of the series
 n = 0
◆ (^) n if it exists.
EXAMPLE 13. 7. Here are a two more examples.
(a) Find the sum of the series
 n = 3
◆ (^) n if it exists.
SOLUTION. First rewrite the series adding back and then subtracting the first few ‘missing’ terms.
 n = 3
◆ (^) n
 n = 0
◆ (^) n!
By Theorem 13. 1 the series converges to (^1) ^1 3
Alternative Method. Write out the first few terms of the series and identify a and r. (^) • Â n = 3
◆ (^) n
|{z}^27 a
|{z}^81 ar
|{z}^243 ar 2
|{z}^729 ar 3
Now a = 278 and the ratio of a term to the previous one is r = 23 and | r | < 1. So by Theorem 13. 1 the series converges to 278 1 ( 23 ) =^
8
(b) Find the sum of the series
 n = 2
◆ (^) n if it exists.
SOLUTION. First rewrite the series; be careful of the signs.
 n = 2
◆ (^) n
 n = 0
◆ (^) n!
By Theorem 13. 1 the series converges to (^1) (^4 1 3 )^
Alternative Method. Write out the first few terms of the series and identify a and r. (^) • Â n = 2
◆ (^) n
|{z}^9 a
|{z}^27 ar
|{z}^81 ar 2
|{z}^243 ar 3
Now a = 49 and the ratio of a term to the previous one is r = 13 and | r | < 1. So by Theorem 13. 1 the series converges to
(^49) 1 ( 13 ) =^
1
(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
Similarly we have Note: We use a horizontal bar to
indicate which part of the decimal
0.6666 · · · = 0.6 = repeats.
and
math 131 infinite series, part vii: absolute and conditional convergence 8
But what about something like 0.12 = 0.121212...? We can write any such expres-
sion as a geometric series. In this case
 n = 0
◆ n
12 100
12 100 99 100
So 0.12 = 0.121212... is rational since it can be written as the fraction 1299.
EXAMPLE 13. 8. Here are a few more repeating decimals to try. Answers: (b) 123999 ; (c) 9999 abcd ; (d) 12229900. (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.
math 131 infinite series, part vii: absolute and conditional convergence 10
Therefore
lim
n! •
a n = lim
n! •
( S n S n 1 ) = lim
n! •
S n lim
n! •
S n 1 = A A = 0,
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 n! • a (^) n 6 = 0, then
 n = 0
a (^) n diverges.
Warning! The nth term test for divergence never allows us to conclude that a
series converges, only that it does not converge. If lim
n! •
a n = 0, we can’t conclude
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
 n = 0
n 2 + 1 3 n 2 + n + 1 converges.
SOLUTION. Notice that
nlim! • a^ n^ =^ nlim! •
n 2 + 1 3 n 2 + n + 1 = (^) nlim! •
1 + (^) n^1 3 + (^1) n + (^1) n
By the nth term test for divergence (Theorem 14. 3 ), the series
 n = 0
n 2 + 1 3 n 2 + n + 1 di- verges.
EXAMPLE 14. 3. Determine whether the series
 n = 1
k n
◆ (^) n converges.
SOLUTION. This time using using one of our key limits (see Theorem 13. 2 )
nlim! • a^ n^ =^ nlim! •
k n
◆ (^) n = e k^6 = 0.
By the nth term test for divergence (Theorem 14. 3 ), the series
 n = 1
k n
◆ (^) n diverges.
EXAMPLE 14. 4. Determine whether the series
 n = 1
p n (^) n converges.
SOLUTION. Using another of our key limits (see Theorem 13. 2 )
nlim! • a^ n^ =^ nlim! •^ n
p n = 1 6 = 0.
By the nth term test for divergence (Theorem 14. 3 ), the series
 n = 1
p n (^) n diverges.
EXAMPLE 14. 5. Determine whether the series
 k = 1
2 n n 2
SOLUTION. Notice that both numerator and denominator both tend to infinity. So converting to x and using l’Hôpital’s rule (twice!) We will need to use the derivative formula (^) dxd ( a x^ ) = a x^ ln a, which is valid when a > 0. nlim! •
2 n n 2 = (^) xlim! •
2 x x 2 = (^) xlim! •
( ln 2 ) 2 x 2 x = (^) xlim! •
( ln 2 )( ln 2 ) 2 x 2
By the nth term test for divergence (Theorem 14. 3 ), the series diverges.
EXAMPLE 14. 6. Determine whether the series
 k = 1
n and
 k = 1
n 2
converge.
SOLUTION. Notice that both lim n! •
n
and lim n! •
n 2
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.
EXAMPLE 14. 7. Determine whether the series
 k = 1
cos (^1) n converges.
SOLUTION. Notice that
nlim! • cos^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.
(a)
 n = 1
3 n + 1 2 n + 5 (b)
 n = 1
n 2 n 2 + 1 (c)
 n = 1
( 1.1 ) n
(d)
 n = 1
n
◆ (^) n (e)
 n = 1
p n (^) n (f )
 n = 1
n! n n
