Theorem: If $\displaystyle{\sum_{n=1}^\infty
a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ are
series with non-negative terms, then:
If $\displaystyle{\sum_{n=1}^\infty b_n}$ converges
and $a_n \le b_n$ for all $n$, then
$\displaystyle{\sum_{n=1}^\infty a_n}$ converges.
If $\displaystyle{\sum_{n=1}^\infty b_n}$ diverges
and $a_n \ge b_n$ for all $n$, then
$\displaystyle{\sum_{n=1}^\infty a_n}$ diverges.
In fact, 1. will work if $a_n\le b_n$ for
all $n$ larger than some finite positive $N$, and
similarly for 2.
Example 1: The series $\displaystyle
\sum_{n=1}^\infty\frac{2^n}{3^n+1}$ converges, since $$
\frac{2^n}{3^n+1}\le \frac{2^n}{3^n} $$ and we know that the
geometric series $\displaystyle
\sum_{n=1}^\infty\left(\frac{2}{3}\right)^n$ is a convergent
geometric series, with $r=\frac23<1$.
The video explains the test, and looks at an example.
Example 2: Test the series
$\displaystyle\sum_{k=1}^\infty\frac{\ln k}{k}$ 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: $\displaystyle\frac{\ln
k}{k}\ge\frac1k$, and the harmonic series with terms $\frac1k$
diverges, so our series diverges.
---------------------------------------------------------------- Example 3: Test the series
$\displaystyle\sum_{n=1}^\infty\frac{1}{5n+10}$ 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
$\displaystyle\frac{1}{5n+10}\le\frac1n$. But the harmonic
series diverges. Our terms are
smaller than those of a divergent series, so we know
nothing. Let's compare to
$\displaystyle\frac1{n^2}$. The series
$\displaystyle\sum\frac{1}{n^2}$ is a convergent $p$-series,
with $p=2$. But when we compare terms, we get
$\displaystyle\frac{1}{5n+10}\ge\frac1{n^2}$ as long as $n\ge7$,
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.