Geometric Riemann Sums
Calum Buchanan
December 2020
1 The problem
We would like to evaluate the definite integral of a continuous function
f
on a closed interval [
a, b
] using Riemann sums. In the canonical example,
we partition [
a, b
] into
k
intervals of equal size and evaluate the sum of the
areas of rectangles of width
b−a
k
and height f(x
i
) where x
i
is some point in
the ith interval. We obtain a sum of the form
k
X
1
f(x
i
)(
b−a
k
),
which is an estimate of the area between the
x
-axis and the curve of
y
=
f
(
x
)
from
x
=
a
to
x
=
b
. The more rectangles we use, the better our estimate
gets, and thus we may evaluate the area under f(x) by taking a limit,
Z
b
a
f(x)dx = lim
k→∞
k
X
i=1
f(x
i
)(
b−a
k
).
Now all we need to do is evaluate this limit. . . . Tricky stuff. We will think
about this problem a little differently using rectangles of varying widths.
2 Geometric series
First, a warm-up. You may have seen a popular proof that 1 = 0
.
999
. . .
. It
goes like this: set
S = 0.999 . . . .
Then,
10S = 9.999 . . . ,
1
and
9S = 10S − S = 9.999 . . . − 0.999 . . . = 9.
Dividing everything by 9, the first and last terms give us
S
= 1. But
S = 0.999 . . ., so 1 = 0.999 . . ..
Since 0
.
999
. . .
=
9
10
+
9
100
+
9
1000
+
· · ·
, we could have phrased the proof
like this: consider the system of equations
10S = 9 +
9
10
+
9
100
+ · · ·
S = 0 +
9
10
+
9
100
+ · · · .
Subtracting the bottom equation from the top equation, we see that 9
S
= 9
and, as before, 0.999 . . . = S = 1.
This is the basic idea of a geometric series. If
q
is some fraction in the
interval (0, 1), then an infinite sum of the form
∞
X
n=1
q
n
= q + q
2
+ q
3
+ q
4
+ · · ·
is an example of a geometric series, and it is always finite (provided
q
stays
strictly between 0 and 1). We can also multiply every term in a geometric
series by a constant, as we did in the previous example. We may have
expressed 0.999 . . . as the geometric series
P
∞
1
9(
1
10
)
n
.
An appropriate next question: Can we always calculate what a geometric
series equals? The answer is yes. Using the same strategy as before, set
S =
P
∞
1
q
n
= q + q
2
+ q
3
+ · · · . Then qS = q
2
+ q
3
+ q
4
+ · · · , and
S − qS = (1 − q)S = q.
Dividing everything by 1
− q
, we obtain
S
=
q
1−q
, which is precisely the
infinite sum
P
∞
1
q
n
. In the previous example, we obtain
S = 9
∞
X
1
1
10
n
= 9 ·
1/10
1 − (1/10)
= 1.
We’ll remember this.
3 Geometric Riemann sums
Say we would like to evaluate the area under the curve
y
=
x
2
on the interval
[0
,
1] using Riemann sums. We can use a geometric series to make our job a
2
(a) q =
1
2
(b) q =
3
4
Figure 1: Geometric Riemann partitions of [0, 1] for the curve y = x
2
.
little easier. Instead of partitioning [0
,
1] into intervals of equal size, we will
partition it into one interval of size
1
2
, one of size
1
4
, one of size
1
8
, etc. as
follows:
1
2
, 1
,
1
4
,
1
2
,
1
8
,
1
4
,
1
16
,
1
8
, . . . .
In other words, we will take intervals of width (
1
2
)
n
for
n
= 1
,
2
,
3
, . . .
. If we
take right-hand Riemann sums, or overestimates, our rectangles will look
like the crude sketch in Subfigure 1a.
Now, this certainly isn’t the best estimate of the area under
x
2
between
x
= 0 and
x
= 1, but we can calculate it using a geometric series. The area
of the blue rectangles,
A
1/2
, is the sum of their heights times their widths, or
A
1/2
= 1 ·
1
2
+
1
2
2
·
1
4
+
1
4
2
·
1
8
+ · · ·
=
∞
X
0
1
2
n
2
·
1
2
n+1
=
∞
X
0
1
2
2n
·
1
2
n+1
=
∞
X
0
1
2
3n+1
.
Adapting the trick from the last section, if
A
1/2
=
1
2
+
1
2
4
+
1
2
7
+
1
2
10
+ · · · ,
then
1
2
3
A
1/2
=
1
2
4
+
1
2
7
+
1
2
10
+ · · · ,
3
and
A
1/2
−
1
8
A
1/2
=
7
8
A
1/2
=
1
2
.
Multiplying every term by
8
7
, we see that
A
1/2
=
1
2
·
8
7
=
4
7
.
As we said, this is a pretty bad estimate of the desired area. However, if
we had instead considered the intervals
3
4
, 1
,
(
3
4
)
2
,
3
4
,
(
3
4
)
3
, (
3
4
)
2
, . . . ,
and had drawn rectangles of width (
3
4
)
n
−
(
3
4
)
n+1
for
n
= 0
,
1
,
2
, . . .
, as in
Subfigure 1b, then the infinite collection of intervals would still cover [0
,
1]
and we could use the same process to obtain an even better estimate. In
fact, we can repeat the above calculations for rectangles of width
q
n
− q
n+1
for any fraction q between 0 and 1 as follows:
A
q
=
∞
X
0
(q
n
)
2
(q
n
− q
n+1
) =
∞
X
0
q
3n
− q
3n+1
= (1 − q)
∞
X
0
q
3n
.
If we set S =
P
∞
0
q
3n
= 1 + q
3
+ q
6
+ · · · , then q
3
S = q
3
+ q
6
+ · · · , and
S − q
3
S = (1 − q
3
)S = 1.
Thus, S =
1
1−q
3
, and
A
q
= (1 − q)S =
1
1 + q + q
2
.
As our fractions
q
become closer to 1, it is not too hard to see that the
approximation of the area between the
x
-axis and the curve
y
=
x
2
improves.
In fact, if we can evaluate the limit of
A
q
as
q
approaches 1, we will obtain
the exact area under the curve. Luckily, this is an easy limit to take, and
Z
1
0
x
2
dx = lim
q→1
A
q
= lim
q→1
1
1 + q + q
2
=
1
3
,
as desired.
Repeat this trick for any function of the form
f
(
x
) =
kx
n
(
k
real and
n
a positive integer) for amazing results.
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References
[1]
George E Andrews. The geometric series in calculus. The American
mathematical monthly, 105(1):36-40, 1998.
[2] Also, thanks to Blair Seidler’s calculus lessons at JHU CTY.
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