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## Finding the Interval of Convergence The main tools for computing the radius of convergence are the
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 $a>0$.) 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. Solution 2: $\displaystyle\left|\frac{\frac{x^{n+1}}{n+1}}{\frac{x^n}{n}}\right|=\left|\frac{x^{n+1}}{n+1}\frac{n}{x^n}\right|=\frac{n}{n+1}|x|\longrightarrow |x|$ |