Determine interval of convergence for $\sum_{n=1}^{\infty} {n!(8x-1)^n}$

Question

We have the infinite series:

$\sum_{n=1}^{\infty} {n!(8x-1)^n}$

First i applied the ratio test giving

$\lim_{x\to \infty} \vert\frac{(n+1)!(8x-1)^{n+1}}{n!(8x-1)^n}\vert$

After simplification:

$\lim_{x\to \infty} {(n+1)(8x-1)}$

${(8x-1)} \cdot \lim_{x\to \infty} {(n+1)}$

The result is we are multiplying our variable ${(8x-1)}$ by $\infty$

What does this tell us about the interval of convergence? I cannot deduce what this is telling me. Did i make a mistake? I would like clarification as to what the conclusion is.

Answer 1

That limit that you got is $\infty$. So, the series always diverges when $x\neq\frac18$. So, the interval of convergence is the singleton $\left\{\frac18\right\}$.