Calculus BC: Series
The Comparison Test
Until the final section of this chapter, we will restrict our attention to series with a n≥ 0 . Thus the partial sums are increasing:
| s 1≤s 2≤ ... ≤s n≤ ... |
If the series
a
n
is to converge, there must be some
B
such that
s
n≤B
for all
n
, or the
s
n
will become arbitrarily large. Such a
B
is called
an upper bound. The value to which the series converges is the least of all possible
upper bounds. It turns out that whenever the sequence
{s
n}
of partial sums has an
upper bound, there exists a least upper bound, to which the series converges. This fact
enables us to prove the comparison test, stated below.
For two series
a
n
,
b
n
, with
a
n, b
n≥ 0
for all
n
, suppose there exists a number
C > 0
such that
| a n≤Cb n |
for all
n
and that
b
n
converges. Then
a
n
converges and
a
n≤C
b
n
|
To prove this statement, it suffices to show that the number
C
b
n
is
a bound for the partial sums
a
1 + a
2 + ... + a
n
. Then the least upper bound of these
partial sums must exist and is clearly less than or equal to
C
b
n
.
Thus we need only note that
| a 1 + ... + a n | ≤ | Cb 1 + ... + Cb n | |
| = | C(b 1 + ... + b n) | ||
| ≤ |
C
b
n
|
A similar test enables us to show that certain series diverge. If
a
n
and
b
n
are again two series with
a
n, b
n≥ 0
for all
n
,
suppose that there exists
C≥ 0
such that
a
n≥Cb
n
for all
n
and that
b
n
diverges. Then
a
n
also diverges. The
proof of this fact is similar to the previous proof--since the partial sums of the
b
n
become arbitrarily large and
| a 1 + ... + a n≥Cb 1 + ... + Cb n = C(b 1 + ... + b n) |
the partial sums of the a n also become arbitrarily large.
