We're looking at
$$
\sum_{n=0}^\infty \frac{(x+3)^n}{n\cdot 3^n}.
$$
Applying the ratio test, we have
$$
\lim_{n\to\infty} \left| \frac{\left(\frac{(x+3)^{n+1}}{(n+1)\cdot 3^{n+1}}\right)}{\left(\frac{(x+3)^n}{n\cdot 3^n}\right)} \right| = \lim_{n\to\infty} \left|\frac{n(x+3)}{3(n+1)}\right| = \lim_{n\to\infty} \left(\frac{|x+3|}{3}\cdot\frac{n}{n+1}\right).
$$
The factor $\dfrac{|x+3|}{3}$ does not change as $n$ changes, so it can be pulled out:
$$
=\frac{|x+3|}{3} \lim_{n\to\infty} \frac{n}{n+1} = \frac{|x+3|}{3}\cdot 1
$$
Thus the series converges if $\dfrac{|x+3|}{3}<1$ and diverges if $\dfrac{|x+3|}{3}>1$.
Now solve the inequality
$$
\frac{|x+3|}{3}<1
$$
for $x$.