Finding the sum of $\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}$ 
Find the sum of the series and for which values of $x$ does it converge: 
$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n} $$

My attempt:
$$\begin{align}&S_n=(x+1)^2+(x+1)^3/3+...+(x+1)^{n+1}/3^{n} \\ 
-(x+1)^2&S_n=-(x+1)^3/3-...-(x+1)^{n+2}/3^{n}  \\ 
&S_n=\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2} \end{align}$$
And for $-4<x<2$ we have:
$$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{(x+1)^2-(x+1)^{n+2}/3^{n}}{1-(x+1)^2}=\frac{(x+1)^2}{1-(x+1)^2}$$
Which is the sum of the series for $-4<x<2$.
The sum looks a bit odd, is that alright? 
Note: no integrals or Taylor or Zeta.

$$\sum_{n=0}^{\infty}\frac {(x+1)^{n+2}}{3^n}=(x+1)^2\sum_{n=0}^{\infty}\Big(\frac {x+1}{3}\Big)^n=(x+1)^2\frac{1}{1-\frac{x+1}{3}}=\frac{3(x+1)^2}{2-x}$$
