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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.

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It is "D'Alembert". – Pedro Tamaroff Apr 1 '13 at 11:56
In fact it is "d'Alembert". – Eckhard Apr 1 '13 at 11:56
@Eckhard The capital D is off? (the e is my usual mistake there) – Pedro Tamaroff Apr 1 '13 at 11:59
@PeterTamaroff I was referring to the 'e'. Also I think, one would usually use a lower-case 'd' except at the beginning of of sentence. – Eckhard Apr 1 '13 at 12:05
A simpler way would be interpreting it as a geometric progression whose convergence criteria is well known ($|r|<1$ where $r$ is the common ratio of the series.) – fermesomme Apr 1 '13 at 12:09

for $|(x+2)/3|<1$ it converges to the limit given by multiplication of geometric series limit and polynomial: $$ \sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x} $$

for $|(x+2)/3|\geq 1$, the sum is not convergent.

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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.

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Hint:$\sum_{k=0}^\infty x^k=\dfrac{1}{1-x}$for $|x|<1$

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