So I started this problem by taking the limit of the power series and applying the root test, giving us $\lim \limits_{x \to \infty}$ $| \frac{(x-6)^n}{(n(-6)^n)}|^\frac{1}{n}$. Canceling out some terms we get the limit of $\frac {x-6}{n(-6)}$. The $x-6$ can be moved to the front of the limit, giving us $ | x-6|$$\lim \limits_{x \to \infty}$ $n(-6)$. The limit is equal to $-\infty$, but around here is where I get lost in terms of finding the interval of convergence. Do we set it up as $|x-6| < 1$ in order to find the radius, or is there another way?
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$\begingroup$Using the root test, we have: $$\lim_{n \to \infty}{|a_n|}^{1/n}=\lim_{n \to \infty}{\frac{|x-6|}{6n^{1/n}}}=\frac{|x-6|}{6}\lim_{n \to \infty}{\frac{1}{n^{1/n}}}=\frac{|x-6|}{6},$$ since $\displaystyle \lim_{n \to \infty}{n^{1/n}}=1.$
Hence, the series converges for $|x-6|<6$, that is, the interval of convergence is $(0,12).$
To see convergence at the endpoints of the interval, you need to test them separately.
When $x=0$, we have: $$\sum_{n=1}^{\infty}{\frac{(-6)^n}{n(-6)^n}}=\sum_{n=1}^{\infty}\frac{1}{n},$$ which diverges (harmonic series).
When $x=12$, we have: $$\sum_{n=1}^{\infty}{\frac{6^n}{n(-6)^n}}=\sum_{n=1}^{\infty}\frac{(-1)^n}{n},$$ which converges (alternating harmonic series).
Therefore, your interval of convergence is $(0,12]$.
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