Theorem: If ∞∑n=1an and ∞∑n=1bn are
series with non-negative terms, then:
If ∞∑n=1bn converges
and an≤bn for all n, then
∞∑n=1an converges.
If ∞∑n=1bn diverges
and an≥bn for all n, then
∞∑n=1an diverges.
In fact, 1. will work if an≤bnfor
all n larger than some finite positive N, and
similarly for 2.
Example 1: The series ∞∑n=12n3n+1 converges, since 2n3n+1≤2n3n
and we know that the
geometric series ∞∑n=1(23)n is a convergent
geometric series, with r=23<1.
The video explains the test, and looks at an example.
Example 2: Test the series
∞∑k=1lnkk for convergence
or divergence.
DO: Do you
think this series converges? Try to figure out what to
compare this series to before reading the solution
Solution 2:lnkk≥1k, and the harmonic series with terms 1k
diverges, so our series diverges.
---------------------------------------------------------------- Example 3: Test the series
∞∑n=115n+10 for convergence
or divergence. DO: Try
this before reading further.
Solution 3: The terms look much like
the harmonic series, and when we compare terms, we see that
15n+10≤1n. But the harmonic
series diverges. Our terms are
smaller than those of a divergent series, so we know
nothing. Let's compare to
1n2. The series
∑1n2 is a convergent p-series,
with p=2. But when we compare terms, we get
15n+10≥1n2 as long as n≥7,
so our terms are larger than those of
a convergent series, and this comparison also tells us
nothing. We will use the limit comparison test (coming
up) to test this series.