Radius of convergence for binomial series (2)

Question

I'm having trouble calculating the radius of convergence for for the following binomial series.

More in detail, I'm having trouble finding $c_k$ and $c_{k+1}$ for the following series:

$$ \sum_{k=0}^{\infty} \binom{3k}{k}x^{2k+1}$$

I'm not sure what to do with the $2k+1$ exponent here.

Answer 1

You have, for each $\require{cancel}k\in\Bbb Z_+$ and each $x\ne0$,\begin{align}\left|\frac{\binom{3(k+1)}{k+1}x^{2(k+1)+1}}{\binom{3k}kx^{2k+1}}\right|&=\frac{\frac{\cancel{(k+2)}\cancel{(k+3)}\ldots(3k+3)}{(2k+2)!}}{\frac{(k+1)\cancel{(k+2)}\cancel{(k+3)}\ldots(3k)}{(2k)!}}|x|^2\\&=\frac{(3k+1)(3k+2)(3k+3)}{(k+1)(2k+1)(2k+2)}|x|^2\\&=3\frac{(3k+1)(3k+2)}{k+1}|x|^2\\&\to\frac{27}4|x|^2\end{align}and therefore the radius of convergence is $\frac2{\sqrt{27}}$.