Sum of $\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n} $ Calculate the sum of the next series and for which values of $x$ it converges:
$$\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n}$$
I used D'Alembert and found that the limit is less than 1, so: $-5 < x < 1$ (because the fraction must be less than 1).
and then I assigned the values: $x=-5$ and $x=1$ in the series and got:
for $x=-5$ and $x=1$, it diverges.
then the series converges in the range of $(-5,1)$, $R=3$ and the center point is for $x=2$.
Please let know if there is a mistake and find the sum.

Your sum is equal to
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n,
$$
which you recognize as a geometric series $\sum_n q^n$ with sum $1/(1-q)$. Thus,
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x}
$$ 
as long as $|(x+2)/3|<1$. 
If $|(x+2)/3|\geq 1$, the sum is not convergent.
