For which values of $x$ does the series converge? $$\sum_{k=0}^{\infty} \frac{3^k\cdot 3^2 \cdot x^k}{2^k}$$
I have to find the values of $x$ for which the series converges.
I used the root test,the limit of the absolute value of the kth root of the series must be smaller than 1 and I ended up with $|3x/2|<1 $ which leads to
 $-2/3< x <2/3 $.
and then I put the once  $-2/3 $ and once  $2/3 $ in the original series and tested for convergence and found that for both values the series diverges which means the series only converges in the open interval  $(-2/3,2/3)$.
Am I right or did I miss something?

The  geometric  series 
\begin{align*}
\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}\qquad\qquad |x|<1
\end{align*}
is convergent for $x\in(-1,1)$.

OPs example is a geometric series (somewhat in disguise), since
  \begin{align*}
\sum_{k=0}^{\infty}\frac{3^k3^2x^k}{2^k}&=3^2\sum_{k=0}^{\infty}\left(\frac{3x}{2}\right)^k\\
&=\frac{9}{1-\frac{3}{2}x}\\
&=\frac{18}{2-3x}
\end{align*}
  which is convergent for 
  \begin{align*}
&\left|\frac{3}{2}x\right|<1\\
\end{align*}
  or equivalently for $x\in\left(-\frac{2}{3},\frac{2}{3}\right)$.

