How do you find the convergence radius of a taylor polynomial
such as $\displaystyle \sum_{n=0}^infty\frac{n^2(x^(2n))}{3^(n-1)}$
Try the ratio test.
$\displaystyle lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n}\right|=L$
if $\displaystyle L < 0$ the series converges
if $\displaystyle L=1$ the test is inconclusive
if $\displaystyle L\ge 1$ The series diverges
$\displaystyle \lim_{n \to \infty}\left| \frac{\frac{(n+1)^2x^{2(n+1)}}{3^{(n+1)-1}}}{\frac{n^2x^{2n}}{3^{n-1}}}\right|$
$\displaystyle \lim_{n \to \infty}\left| \frac{(n+1)^2x^{2n+2}}{3^{n}} \cdot \frac{3^{n-1}}{n^2x^{2n}}\right|$
$\displaystyle \lim_{n \to \infty}\left| \frac{(n+1)^2x^2}{3n^2}\right|=\left|\frac{x^2}{3} \right|$
So we want the series to converge so we want
$\displaystyle \left| \frac{x^2}{3}\right| < 1$
if we solve the inequality we get the radius of convergance.
Note you need to check the end points with another test.