Find the radius of convergence of R of the following power series.

$\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$

I don't know what to do with the $\displaystyle x^{2k}$ term.

Thanks

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- Dec 10th 2007, 08:02 PMtbyou87Power Series Convergence Radius
Find the radius of convergence of R of the following power series.

$\displaystyle \sum_{k=0}^{\infty}\frac{k2^k}{3^k}x^{2k}$

I don't know what to do with the $\displaystyle x^{2k}$ term.

Thanks - Dec 10th 2007, 08:18 PMJhevon
- Dec 10th 2007, 08:20 PMThePerfectHacker
Use the root test,

$\displaystyle \left| \frac{k2^k}{3^k}x^{2k} \right|^{1/k} = \frac{2}{3}k^{1/k}x^2$ the limit of $\displaystyle k^{1/k}\to 1$ thus:

$\displaystyle |a_kx^{2k}|^{1/k} \to \frac{2}{3}x^2$.

We require that,

$\displaystyle \frac{2}{3}x^2 < 1 \implies x^2 < \frac{3}{2}$.