Find the convergence domain of the series $\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$ 
Exercise: Find the convergence domain of the series $$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}$$

To solve the series I used the comparison test then the Leibniz criteria:
$$\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^n(x-5)^n}\leqslant\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{(x-5)^n}$$
Now applying Leibniz criteria the series converge for $x\neq 5$.
Solution(book): $x\geqslant5\frac{1}{3},x<4\frac{2}{3}$.
Question:
The solution surprised me. What am I doing wrong? Why is it not the same?
Thanks in advance!

The OP has asked, in comments beneath various answers as well as in the question itself, what's wrong with his procedure, so I'm going to address that only.
There seem to be two fundamental misconceptions.  First, in the comparison step, it looks like you are saying that 
$${1\over n3^n}\le 1\implies {(-1)^{n-1}\over n3^n(x-5)^n}\le{(-1)^{n-1}\over(x-5)^n}$$
But that implication does not hold when ${(-1)^{n-1}\over(x-5)^n}$ is negative.
Second, in the "Leibniz criteria" (aka, Alternating Series Test) step, you seem to be assuming that as long as $x\not=5$, the sequence $1\over(x-5)^n$ decreases monotonically to $0$ (that's what the Alternating Series Test requires in order to conclude convegence).  But this is only true if $x\gt6$. If $x\lt5$, the sequence is alternating, so it doesn't decrease, if $5\lt x\lt 6$, the sequence actually increases to $\infty$, and if $x=6$, it's the constant sequence $1$, so it doesn't decrease to $0$.
