Find a power series representation for $\frac{1}{2+x}$ 
Find a power series representation for $\frac{1}{2+x}$

The question did not specify a real number $c$ and an expansion on the form
$$f(x)=\sum_{n\geq 0} a_n(x-c)^n.$$
I did this problem as follows, but I seem to be off by a factor of $1/3$.
$$\frac{1}{2+x} = \frac{1}{1+(x+1)}=\frac{1}{1-[-(x+1)]}.$$
Putting this into the geometric form of:
$$\sum_{n=0}^{\infty} \frac{a}{1-r},$$ I get $a=1$ and $r=-(x-1).$ So, this could be written as $$\sum_{n=0}^{\infty} [(-1)(x-1)]^n = \sum_{n=0}^{\infty} (-1)^n(x-1)^n.$$
Wolfram's widget gets the following:
$$\frac{1}{2+x}=\sum^\infty_{n=0}(-1+x)^n(-1)^n3^{-1-n}\quad \text{for}|-1+x|\lt3$$
which looks like my answer except for the $3^\left(-1-n\right)$ term.
Can you assist me in finding my error?  I don's see where the $1/3$ term gets introduced.
 A: Here is another answer: For this rational function it is possible to give an elementary construction of a powerseries and an "equality"
$$ \frac{1}{2+x}=\sum_{n \geq 0} a_n(x-a)^n$$
for any $a\neq -2$:
$$\frac{1}{2+x}=\frac{1}{2(1-(-x/2))} =\frac{1}{2}\sum_{n\geq 0}(-\frac{x}{2})^n=$$
$$ \sum_{n\geq 0}(-1)^n\frac{x^n}{2^{n+1}}.$$
I used
$$\frac{1}{1-t}=\sum_{n\geq 0} t^n$$
with $t:=-\frac{x}{2}$.
We can generalize as follows (let $a\neq -2$):
$$ \frac{1}{2+x}=\frac{1}{2+a-(-(x-a))} =$$
$$ \frac{1}{2+a}\frac{1}{1-(-(\frac{x-a}{2+a}))}=$$
(here we let $t:=\frac{x-a}{2+a}$)
$$ \frac{1}{2+a}\sum_{n\geq 0}(-(\frac{x-a}{2+a}))^n.$$
We get a power series
$$\frac{1}{2+x}= \frac{1}{2+a}\sum_{n\geq 0}(-1)^n(\frac{x-a}{2+a})^n.$$
With $a:=1$ we get
$$\frac{1}{2+x}=\frac{1}{3}\sum_{n \geq 0}(-1)^n(\frac{x-1}{3})^n.$$
When I write $=$ in the above equation this means for all $x$ where the powerseries converge. You should be able to calculate this radius of convergence yourself.
In general: If $\frac{f(x)}{g(x)}$ is a rational function over the reals, there is a product decomposition
$$P1.\text{  }g(x)=\prod_{i} (x-c_i)^{l_i}\prod_j p_j(x)^{k_j}$$
where $c_i$ are real numbers and $p_j(x)=b_j^2+(x-a_j)^2$ where $a_j,b_j\neq 0$ are real numbers. For a rational function on the form $\frac{1}{p_j(x)}$
you get a similar calculation:
$$ \frac{1}{p_j(x)}=\frac{1}{b_j^2+(x-a_j)^2}=$$
$$\frac{1}{p_j(x)}"="\frac{1}{b_j^2}\sum_{n\geq 0}(-1)^n(\frac{x-a_j}{b_j})^{2n} = \sum_{n \geq 0} \frac{(-1)^n}{b_j^{2(n+1)}}(x-a_j)^{2n}.$$
Hence you can "in principle" construct an explicit formal power series expansion for any rational function $\frac{f(x)}{g(x)}$ using the geometric series and partial fraction decomposition of the product in P1.
A: The confusion comes from the fact that your answer is wrong. See the answer from Z Ahmed for a correct solution. In your solution you somehow transformed $x+1$ into $-x+1$. This happened when you wrote $r$.
Note that the different answers write a series around different points:

*

*yours is around $-1$, with small $x+1$

*the given answer, and the one from ZA is around $1$, with small $x-1$

*hm2020 expands around $0$, with small $x$
A: $$f(x)=\frac{1}{2+x}=\frac{1}{3+(x-1)}=\frac{1}{3}~~\frac{1}{[1+(x-1)/3]}$$
Use IGP
$$\implies f(x)=\frac{1}{3}\sum_{k=0}^{\infty} (-1)^k [(x-1)/3] ^k, ~if~ |x-1|<3. $$
