Download Arithmetic and Geometric Series I. Arithmetic Series

Arithmetic and Geometric Series
Dave L. Renfro
Central Michigan University
November 8, 2002
I. Arithmetic Series
1. Evaluating 1 + 2 + 3 +
+ 100
Let S represent this sum. Here’s a neat way to evaluate S. First, add S to itself in the
following way:
S
S
=
=
1
100
+
+
S+S =
2
99
+
+
+
3
98
+
+
+
+
+
+
+
98
3
99
2
+
On the right hand side of the last equation we have
times. Therefore, it follows that 2S = (
)(
equation by 2 to get S:
S =
:
2. Evaluating 1 + 2 + 3 +
+
+
+
+
100
1
+
added to itself
many
). Now divide both sides of this
+n
Let T (n) represent this sum. Then T (1) =
, T (3) =
, T (7) =
,
1
and T (100) =
. By doing the same thing that we did in Section 1 above, we can
get a closed form expression for T (n).
T (n)
T (n)
=
=
T (n) + T (n) =
1
n
+
2
+
3
+
+ (n 1) + (n 2) +
+ (n 2) + (n 1) +
+
3
+
2
+
+
+
+
+
On the right hand side of the last equation we have
times. Therefore, it follows that 2T (n) = (
)(
equation by 2 to get T (n):
T (n) =
1
+
n
1
+
added to itself
many
). Now divide both sides of this
:
The numbers T (1), T (2), etc. are called triangular numbers. For the reason why they are called this, look
at some of the web pages you’ll …nd at <http://www.google.com/search?q=triangular+numbers>.
3. Numerical examples using 1 + 2 + 3 +
Example 1: 3 + 6 + 9 +
=
3(
+ 1)
+ 663
+
+
+
Example 2: 4 + 7 + 10 +
+
(
+ 3) + (
+ 6) + (
=
(
+
+
+
=
(
) (
)
+
)
3T (
)
3T (
)
=
:
(
+ 5) + (
+ 10) + (
=
(
+
+
3(
+ 663)
+
=
+
+
+
)
:
=
(
+ 15) + (
+
+
+
+ (
+ 1257
=
)
=
+ 9) +
+
Example 3: 12 + 17 + 22 + 27 +
+
)
+ 664
=
) (
1
n (n
2
+n =
)
(
+
(
+ 20) +
) (
) T(
)
4. Algebraic examples using 1 + 2 + 3 +
+
+
=
+
+
)
+
)
:
1
n (n
2
+n =
+ 1)
Example 1: 1 + 2 + 3 +
+ (n
1)
=
T(
)
=
.
Example 2: 1 + 2 + 3 +
+ (n + 3) + (n + 4)
=
T(
)
=
.
Example 3: 1 + 2 + 3 +
2) + (n
+ (
+ (2n
1) + 2n
=
T(
)
=
.
5. An explicit formula for the sum of an arithmetic series
Suppose the …rst term of an arithmetic series is a, its common di¤erence is d, and the number
of terms is n. Use a, d, and n to express the terms of this series in the manner indicated:
Sn
=
(1’st term) + (2’nd term) + (3’rd term) +
Sn
=
(
) + (
) + (
+ (n’th term)
) +
+ [
]
If we group the a’s together, then factor a d out of the remaining terms, and …nally use an
appropriate evaluation of the function T , we can …nd a closed form expression for the sum
in the same way that the numerical examples 2 and 3 were done above.
Sn
=
Sn
=
(
(
+
) (
+
+
)
+
+
(
)
+
) T(
(
) (
)
=
+
+
+
:
+
)
Now show that Sn can be written in the following two ways:
Sn
n
2
=
[2a + (n
Sn
1) d]
n
2
=
(a1 + an )
6. How to recognize an arithmetic series
A series is an arithmetic series if (a) consecutive terms di¤er by the same amount, or
N
P
(b) in sigma notation the series has the form
an , where an is a linear function of n.
n=M
II. Geometric Series
1. Prelude #1: Find a fraction for 0:37373737:::
You may have seen this method in a beginning algebra class. First, set the repeating decimal
equal to x and multiply both sides of this equation by 10n , where n is the period length of
the repeating decimal. Then subtract the original equation from this new equation and solve
for x. For the repeating decimal 0:37373737::: the period length is 2.
x = 0:37373737:::
100x = 37:37373737:::
100x
99x
x
x = (37 + 0:37373737:::)
= 37
= 37=99
(0:37373737:::)
2. Prelude #2: Find an explicit value for the in…nite sum
4
3
+
8
9
+
16
27
+
Set the sum equal to x and multiply both sides of this equation by 32 , which happens to
be the ratio between consecutive terms. Then subtract the original equation from this new
equation and solve for x.
x = 43 + 89 + 16
27 +
2
3x
2
3x
1
3x
x
x =
=
= 4
8
9
4
3
+
16
27
=
8
9
+
+
32
81
+
16
27
+
32
81
+
4
3
+
8
9
+
16
27
+
3. How these two examples are similar
37
37
37
Note that 0:37373737::: = 100
+ 10000
+ 1000000
+
. Thus, the …rst example above
can be viewed as an in…nite geometric series. Moreover, the method that we used in the
…rst example to …nd a fraction equivalent to 0:37373737::: is essentially same method that
we used in the second example. To see this, note that in the …rst example the common ratio
1
was 100
. Although we actually multiplied the original equation in the …rst example by 100,
1
1
, multiplying by 100
would have accomplish the same thing, namely shifting
and not 100
the decimals over by two spaces to allow almost everything to line up and cancel.
x = 0:37373737:::
= 0:00373737:::
1
100 x
1
100 x
99
100 x
x
x
=
(0:00373737:::)
=
0:37
=
37=99
=
(0:37 + 0:00373737:::)
37
100
4. The sum of an in…nite geometric series
Consider an in…nite geometric series with common ratio r, where 1 < r < 1. [We’ll see later
why we need this restriction on r.] If a is the …rst term, then we’re looking at a + ar + ar 2 +
.
Set the sum equal to x and multiply both sides of this equation by r. Then subtract the
original equation from the new equation and solve for x.
x =
a + ar + ar2 +
rx = ar + ar2 + ar3 +
rx
(r
x
x
= ar + ar2 + ar3 +
1) x =
a
= a = (1 r)
a + ar + ar2 +
Note if a = 1 and r = 2, then this formula says that 1 + 2 + 4 + 8 +
equals 1! More
generally, we’ll get a negative number if a is positive and r > 1. This much alone tells us
that we don’t want to use this formula when r > 1.
5. The sum of a …nite geometric series
The same technique used above also works for a …nite geometric series. Suppose the …rst
term of a …nite geometric sequence is a, its common ratio is r (for a …nite series no restriction
on r is needed), and the number of terms is n. Use a, r, and n to express the sum of the
terms of this sequence in the manner indicated:
rSn
Sn
=
(1’st term) + (2’nd term) + (3’rd term) +
Sn
=
(
) + (
) + (
) +
+ (
)
rSn
=
(
) + (
) + (
) +
+ (
)
Sn = (
(r
1) Sn =
Sn
=
)
+ (n’th term)
(
)
Now show that Sn can be written in the following way:
Sn =
a(1
1
rn )
r
6. Two ways to verify the formula we just found
One way to show that
a (rn 1)
r 1
=
a + ar + ar2 +
+ arn
1
is to simply verify by multiplication that
a (rn
1)
=
(r
+ arn
1) a + ar + ar2 +
1
:
Expanding the right hand side gives
ar + ar2 + ar3 +
a
ar
ar2
Combining like terms leaves us with
ar3
+ arn
1
arn
1
+ arn
a + arn = a ( 1 + rn ) = a (rn
1).
Another way to verify this result is to use synthetic division.2 Since we can cancel the factors
of a, it will be enough to show that
rn
r
1
1
1 + r + r2 +
=
+ rn
1
+ xn
1
:
Note this is the same thing as showing that
xn
x
1
1
1 + x + x2 +
=
:
Here’s a synthetic division table for this division:
xn
1c
xn
1
x
xn
0
1
1
n 1
1
1
n 2
x
2
x2
x
0
1
0
1
0
1
1
1
1
1
1
0
x
c o n sta nt
re m a in d e r
n 3
x
c o n sta nt
7. Convergence of in…nite geometric series
If
1 < r < 1, then rn ! 0 when n ! 1. Therefore, for these values of r we have
a + ar + ar2 +
+ arn
1
+
!
a (0 1)
a
a
=
=
;
r 1
r 1
1 r
which is the same formula that we got back in Section 4.
However, we will not assign a numerical sum to an in…nite geometric series if jrj 1, at least
when a 6= 0.3 This is because as you add more and more terms, the partial sums don’t get
closer and closer to any number when r
1 or r 1. In fact, the terms you are adding
will get “in…nitely large”(either positively or negatively) if r < 1 or r > 1, and you should
be able to verify for yourself what happens when r = 1 or r = 1.
8. How to recognize a geometric series
A series is a geometric series if (a) consecutive terms have the same ratio, or (b) in sigma
N
P
notation the series has the form
an , where an is an exponential function of n.
n=M
2
Incidentally, mathematical induction can be used to prove that rn 1 is divisible by r 1 for each n = 1,
2, 3, ..., where the identity rn+1 1 = (rn 1) r + (r 1) is used for the inductive step. However, this
method will only show that rn 1 is divisible by r 1. It won’t tell you what the quotient will be.
3
Obviously, if a = 0, then it is reasonable to say that the in…nite sum a + ar + ar2 +
is zero.