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## The Ratio Test
Let $\sum a_n$ be a series. The
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 simiplificationAs 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. DO: Simplify
$\displaystyle\frac{\frac{50^{n+1}}{(n+1)!}}{\frac{50^n}{n!}}$$\displaystyle\frac{\frac{50^{n+1}}{(n+1)!}}{\frac{50^n}{n!}}=\frac{50^{n+1}}{(n+1)!}\cdot\frac{n!}{50^n}=\frac{50^{n+1}}{50^n}\cdot\frac{n!}{(n+1)!}=50\cdot\frac{1}{n+1}=\frac{50}{n+1}$ A couple of worked out examples of the Ratio Test are contained in the video, as well as the ideas of why the Ratio Test works. |