How to prove that sum given by generating function diverges for given value of $x$

Question

I have a generating function: $A(x)=\dfrac{3-8x}{1-4x+6x^2-3x^3}$ (also I have a recurrence from which this function is built). I have to prove that sum $\sum\limits_k a_k\left(\dfrac{4}{3}\right)^k$ diverges, where $a_k$ is $k$-th recurrence member.

As far as I found in books standard method for getting asymptotic for $a_k$ is to factor $A(x)$ into simple fractions. But in this case I have complex roots for cubic equation and some unusable coefficients in $\dfrac{A}{x_1-x}+\dfrac{B}{x_2-x}+\dfrac{C}{x_3-x}$. So I wonder if there is some simpler/other method for getting asymptotic or proving divergence.

I've looked on something about 150 members of recurrence using WolframAlpha: members change sign (six members negative, six positive, and so on) and absolute value increases. But this method hardly counts as a proof.

Thanks in advance for any ideas.

Answer 1

Hint: The given function has a singularity at $x=1$, so its radius of convergence cannot be larger than $1$.