What is the radius of convergence of the power-series \(\displaystyle \sum_{n=0}^\infty \frac{(x-4)^{2n}}{3^n}\)  ?

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Using the ratio test:

$$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{(x-4)^{2n+2}}{3^{n+1}}\cdot \frac{3^n}{(x-4)^{2n}} $$ $$ = \lim_{n\to\infty}\frac{(x-4)^2}{3} $$

The power-series converges when the limit is less than 1.

$$ \frac{(x-4)^2}{3} \lt 1 $$ $$ (x-4)^2 \lt 3 $$ $$ x-4 \lt \pm\sqrt{3} $$