My calc book is very vague

**Mathstud28** · April 2nd 2008, 01:40 PM

How do you find the convergence radius of a taylor polynomial

such as $\sum_{n=0}^infty\frac{n^2(x^(2n))}{3^(n-1)}$

**Mathstud28** · April 2nd 2008, 01:52 PM

$\sum_{n=0}^{\infty}\frac{n^2x^{2n}}{3^{n-1}}\$

**TheEmptySet** · April 2nd 2008, 03:10 PM

Originally Posted by Mathstud28

$\sum_{n=0}^{\infty}\frac{n^2x^{2n}}{3^{n-1}}\$

Try the ratio test.

$lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n}\right|=L$

if $L < 0$ the series converges
if $L=1$ the test is inconclusive
if $L\ge 1$ The series diverges

$\lim_{n \to \infty}\left| \frac{\frac{(n+1)^2x^{2(n+1)}}{3^{(n+1)-1}}}{\frac{n^2x^{2n}}{3^{n-1}}}\right|$

$\lim_{n \to \infty}\left| \frac{(n+1)^2x^{2n+2}}{3^{n}} \cdot \frac{3^{n-1}}{n^2x^{2n}}\right|$

$\lim_{n \to \infty}\left| \frac{(n+1)^2x^2}{3n^2}\right|=\left|\frac{x^2}{3} \right|$

So we want the series to converge so we want

$\left| \frac{x^2}{3}\right| < 1$

if we solve the inequality we get the radius of convergance.

Note you need to check the end points with another test.

**Jhevon** · April 2nd 2008, 04:25 PM

Originally Posted by Mathstud28

$\sum_{n=0}^{\infty}\frac{n^2x^{2n}}{3^{n-1}}\$

Originally Posted by Mathstud28

**Mathstud28** · April 2nd 2008, 04:27 PM

Obviously we want $a_{n+1}\$ < $a_{n}\$ because then that is what causes convergence?

**Jhevon** · April 2nd 2008, 04:31 PM

Originally Posted by Mathstud28

Obviously we want $a_{n+1}\$ < $a_{n}\$ because then that is what causes convergence?

no, we want $\lim_{n \to \infty} \left| \frac {a_{n + 1}}{a_n} \right| < 1$ ............then we have convergence by the ratio test

all you need to do is solve the last line TheEmptySet left you with for x. be sure to check the end-points as well

**Mathstud28** · April 2nd 2008, 05:20 PM

if its not too much to ask could you give me an example I work through it and you see if I am right? If thats too much work I understand...with all that LaTeX you have to go through to do it I wouldn't blame you haha

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