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!

We have
$$
\sum_{k=0}^{\infty}(-1)^k y^k = \lim_{n\rightarrow \infty}\frac{y^n+1}{y+1}
$$
here $\vert y\vert < 1 \Rightarrow$ convergence
Now
$$
\sum_{k=0}^{\infty}(-1)^k \frac{y^k}{k} = \frac{1}{y}\int_0^y \sum_{k=0}^{\infty}(-1)^k y^k dy = \frac{1}{y}\int_0^y\frac{dy}{y+1} = \frac{\ln(y+1)}{y}
$$
for $y \ne 0$ and $\vert y\vert< 1$ so the convergence is linked to
$\vert\frac{1}{3(x-x)}\vert < 1 \Rightarrow 5\frac{1}{3} < x$ and $x < 4\frac{2}{3}$
