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 $(-1,1]$.