Let $\sum a_n$ be a series. The Ratio
Test involves looking at $$\displaystyle{\lim_{n
\to \infty} \frac{\left|a_{n+1}\right|}{\left|a_n\right|}}$$ to
see how a series behaves in the long run. As $n$ goes to
infinity, this ratio measures how much smaller the value of
$a_{n+1}$ is, as compared to the previous term $a_n$, to see how
much the terms are decreasing (in absolute value). If this
limit is greater than 1, then for all values of $n$ past a certain
point, the ratio
$\frac{\left|a_{n+1}\right|}{\left|a_n\right|}>1$, which would
indicate that the series is no longer decreasing. On the
other hand, if this limit is less than 1, the series converges
absolutely.
The Ratio Test: Suppose that
$$\displaystyle{\lim_{n\to\infty}
\frac{\left|a_{n+1}\right|}{\left|a_n\right|}} = L.$$
If $L < 1$, then $\sum a_n$ converges absolutely.
If $L > 1$, or the limit goes to $\infty$, then
$\sum a_n$ diverges.
If $L=1$ or if $L$ does not exist, then this test is
inconclusive, and we must do more work. We say the
Ratio Test fails if $L=1$
Notice that the Ratio Test considers the ratio
of the absolute values of the terms. As you
might expect, the Ratio Test thus gives us information about whether
the series $\sum a_n$ converges absolutely.
Warning: There are
examples with $L=1$ that converge absolutely, examples that converge
conditionally, and examples that diverge. DO: Apply the Ratio Test to 1) the
absolutely convergent series $\sum\frac{1}{n^2}$, 2) the
conditionally convergent series $\sum-\frac{1}{n}$, and 3) the
divergent series $\sum\frac{1}{n}$.
1) As $n\to\infty$,
$\displaystyle\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}}=\frac{n^2}{(n+1)^2}\longrightarrow
1$.
2) As $n\to\infty$,
$\displaystyle\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{1}{n+1}}{\frac{1}{n}}=\frac{n}{n+1}\longrightarrow
1$.
3) As $n\to\infty$,
$\displaystyle\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{1}{n+1}}{\frac{1}{n}}=\frac{n}{n+1}\longrightarrow
1$.
We used other tests to determine the convergence/divergence of these
series - the Ratio Test fails
to help us with these series.
DO: The only way a series could
be conditionally convergent is if the Ratio Test fails for that
series. Why?
Review of simiplification
As you work through this module, you must be able to work with
ratios of factorials as well as ratio of powers. Recall that
$n!=1\cdot 2\cdot 3\cdots(n-1)\cdot n$. DO:
Simplify $\frac{(n+1)!}{n!}$.
$\frac{(n+1)!}{n!}=\frac{1\cdot 2\cdot 3\cdots(n-1)\cdot n\cdot
(n+1)}{1\cdot 2\cdot 3\cdots(n-1)\cdot n}=n+1$ after all the
cancellation.