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 $x$.