I know that the first step would be to use the ratio test. You would add a $+1$ to everywhere there is an n and then multiply that by the reciprocal of the original function but what would I do next after that?
$$\sum_{n=0}^\infty \frac{(x-2)^n}{4^n\sqrt{n}}$$
How do I simplify the double fraction result after I set up the ratio test?
Note that $n$ should start at a number that's greater than zero because, as you probably know, you can't have zero in the denominator.
$$ \lim_{n\to\infty}\left|\frac{\left(\frac{(x-2)^{n+1}}{4^{n+1}\sqrt{n+1}}\right)}{\left(\frac{(x-2)^n}{4^n\sqrt{n}}\right)}\right|= \lim_{n\to\infty}\left|\frac{(x-2)\sqrt{n}}{4\sqrt{n+1}}\right|= \left|\frac{x-2}{4}\right|\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}=\\ \left|\frac{x-2}{4}\right|\lim_{n\to\infty}\sqrt{1-\frac{1}{n+1}}= \left|\frac{x-2}{4}\right|\cdot \sqrt{1-0}= \left|\frac{x-2}{4}\right|. $$
So, the series $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{4^n\sqrt{n}}$ converges absolutely when: $$ \left|\frac{x-2}{4}\right|<1\implies\\ |x-2|<4\implies -2<x<6. $$
And the last thing you need to do is to figure out if the series converges at the endpoints $-2$ and $6$ because if $\left|\frac{x-2}{4}\right|=1$ (that happens precisely when $x=-2$ or $6$), the ratio test is inconclusive. If $x=-2$, the seires $\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{1}{\sqrt{n}}$ converges by the alternating series test and if $x=6$, the series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ diverges by the p-series test because in this case $p=\frac{1}{2}<1$.
Therefore, the interval of convergence of the series $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^n}{4^n\sqrt{n}}$ is $[-2,6)$ and its radius of convergence is $4$.