The Limit Comparison Test
is a good test to try when a basic comparison does not work (as in
Example 3 on the previous slide). The idea of this test is
that if the limit of a ratio of sequences is 0, then the
denominator grew much faster than the numerator. If the
limit is infinity, the numerator grew much faster. If your
limit is non-zero and finite, the sequences behave similarly so
their series will behave similarly as well.
Limit Comparison Test: Let
∞∑n=1an and
∞∑n=1bn be positive-termed
series. If limn→∞anbn=c,
where c is finite, and c>0,
then either both series converge or both diverge.
Example 1 (from previous page): We were trying to
determine whether ∞∑n=115n+10
converges or diverges, and the basic comparison test was not
helpful. DO: Try the limit
comparison test on this series, comparing it to the harmonic
series, before reading further.
Solution 1: It does not matter
which series you choose to have terms an and bn. anbn=15n+101n=15n+10n1=n5n+10.
Then
limn→∞anbn=limn→∞n5n+10=15
and 0<15<∞, so the series behave the same.
Since the harmonic series diverges, so does our series. DO: What if we had chosen an and
bn the other way around? Can you see why it doesn't
matter?
Example 2: To determine whether the series
∞∑n=14n2n+3n converges or
diverges, we'll look for a series that "behaves like" it when n is
large.
Solution 2: Since we think 4n2n+3n≈4n3n,
when n is large,
we'll use ∞∑n=1(43)n for
comparison. (DO: Why can we
not use the basic comparison test with this series?)
Since limn→∞4n2n+3n4n3n=limn→∞3n2n+3n=limn→∞3n2n+3n13n13n=limn→∞1(2n3n)+1=1,
and
0<1<∞, our series are comparable. Since the
geometric series
∞∑n=1(43)n diverges
(r=43>1), we can conclude that our original series
diverges as well.
The following test, which was discussed in the
video, is not explicitly used by most instructors, and is not in
most calculus texts -- discuss it with
your instructor before using.
Theorem:
Suppose that ∞∑n=1an and
∞∑n=1bn are positive-termed
series, c is a positive constant, and that N is some
positive integer.
If ∞∑n=1bn converges
and an≤cbn for all n>N, then
∞∑n=1an converges.
If ∞∑n=1bn diverges
and an≥cbn for all n>N, then
∞∑n=1an diverges.