Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?

What about the ratio test?
\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\to\infty}\frac{n3^{n}|x-5|^{n}}{(n+1)3^{n+1}|x-5|^{n+1}}\\\\
& = \limsup_{n\to\infty}\left(\frac{n}{n+1}\right)\frac{1}{3|x-5|} = \frac{1}{3|x-5|} < 1\\\\
& \Rightarrow |x - 5| > \frac{1}{3} \Rightarrow \left(x > \frac{16}{3}\right)\vee\left(x < \frac{14}{3}\right)
\end{align*}
Moreover, for $x = 16/3$, one has that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}3^{-n}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < \infty
\end{align*}
due to the Leibniz test.
On the other hand, for $x = 14/3$, we have that
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}(-1)^{n}3^{-n}} = -\sum_{n=1}^{\infty}\frac{1}{n} = \infty
\end{align*}
since it is the harmonic series.
Hopefully this helps.
