### Finding the Interval of Convergence

The main tools for computing the radius of convergence are the Ratio Test and the Root Test.  To see why these tests are nice, let's look at the Ratio Test.  Consider $\displaystyle\sum_{n=1}^\infty c_nx^n$, and let $\lim\left|\frac{c_{n+1}}{c_n}\right|=L$.  The Ratio Text will look at $$\displaystyle\lim_{n\to\infty}\left|\frac{c_{n+1}x^{n+1}}{c_nx^n}\right|=\lim_{n\to\infty}\left|\frac{c_{n+1}}{c_n}\right||x|=L|x|,$$and by the Ratio Test the series converges absolutely for $L|x|<1$, i.e. for $|x|<\frac{1}{L}$.  So the radius of convergence is $\frac{1}{L}$ and the interval of convergence is from $-\frac{1}{L}$ to $\frac{1}{L}$.  The same logic holds for the Root Test.  The endpoints of the interval of convergence must be checked separately, as the Root and Ratio Tests are inconclusive there (when $x=\pm\frac{1}{L}$, the limit is 1).  To check convergence at the endpoints, we put each endpoint in for $x$, giving us a normal series (no longer a power series) to consider.  All the tests we have been learning for convergence can be used to test for convergence at the endpoints:  the Divervence Test, $p$-series, Alternating Series Test, Comparison Test, Limit Comparison Test, and/or the Integral Test.

In the video, he uses $a_n$ instead of $c_n$ for the power series coefficients to explain the paragraph above.  However, you do not need to memorize that $R=\frac{1}{L}$, for example -- just use the Ratio or Root Test and follow it to its conclusion.

DO:  work the following without looking at the solutions, which are below the examples.

Example 1:  Find the radius of converge, then the interval of convergence, for $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n^2x^n}{2^n}$.

Example 2:  Find the radius of converge, then the interval of convergence, for $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{x^n}{n}$.

Solution 1:
$\displaystyle\sqrt[n]{\left|\frac{n^2x^n}{2^n}\right|}=\sqrt[n]{n^2}\frac{|x|}{2}\longrightarrow\frac{1}{2}\vert x\vert\quad$ (We used our very handy previous result:  $\sqrt[n]{n^a}\rightarrow 1$ for any $a>0$.)

By the Root Test, our series converges when $\frac{1}{2}\vert x\vert<1$, i.e. when $|x|<2$, so $R=2$.  Now we check the endpoints of the interval from $-2$ to $2$.
When $x=-2$, we have $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n^2x^n}{2^n}=\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n^2(-2)^n}{2^n}=\displaystyle\sum_{n=1}^\infty\frac{n^2(2)^n}{2^n}=\displaystyle\sum_{n=1}^\infty n^2$, which diverges by the Divergence Test.
When $x=2$, we have $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n^2x^n}{2^n}=\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n^2(2)^n}{2^n}=\displaystyle\sum_{n=1}^\infty(-1)^n n^2$ which also diverges by the Divergence Test.

$R=2$ and our interval of convergence is $(-2,2)$.
Solution 2:
$\displaystyle\left|\frac{\frac{x^{n+1}}{n+1}}{\frac{x^n}{n}}\right|=\left|\frac{x^{n+1}}{n+1}\frac{n}{x^n}\right|=\frac{n}{n+1}|x|\longrightarrow |x|$

By the Ratio Test, our series converges when $|x|<1$, so $R=1$.  We test at our endpoints of the interval from $-1$ to $1$:
When $x=-1$, we have $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{x^n}{n}=\displaystyle\sum_{n=1}^\infty(-1)^n\frac{(-1)^n}{n}=\displaystyle\sum_{n=1}^\infty\frac{1}{n}$ which is the divergent harmonic series.

When $x=1$, we have $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{x^n}{n}=\displaystyle\sum_{n=1}^\infty(-1)^n\frac{1^n}{n}\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n}$ which is the convergent alternating harmonic series (use the AST if you are uncertain).

$R=1$ and our interval is $(-1,1]$.