Power Series Representation of $(1+x)/(1-x)$ For the power series representation of, $f(x) = \frac{1+x}{1-x}$ which is $1 + 2 \sum_{n=1}^\infty x^n$, Where does the added $1$ in front come from? How do I get to this answer from $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1}$
 A: $$f(x)=\frac{1+x}{1-x}=(1+x)\frac{1}{1-x}$$
$$=(1+x)\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}x^n+x\sum_{n=0}^{\infty}x^n=$$
$$=\sum_{n=0}^{\infty}x^n+\sum_{n=0}^{\infty}x^{n+1}$$
because 
$$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=1}^{\infty}x^n$$
and
$$\sum_{n=0}^{\infty}x^{n+1}=\sum_{n=1}^{\infty}x^{n}$$
we get
$$(1+x)\frac{1}{1-x}=1+\sum_{n=1}^{\infty}x^n+\sum_{n=1}^{\infty}x^n=$$
$$=1+2\sum_{n=1}^{\infty}x^n$$
A: The first sum is 
$$\sum\limits_{n=0}^\infty x^{n}=1+\sum\limits_{n=1}^\infty x^{n}.$$ By changing the summation index $k=n+1$ the second sum can be rewritten as $$\sum\limits_{n=0}^\infty x^{n+1}=\sum\limits_{k=1}^\infty x^{k}$$ thus
 $$\displaystyle\sum\limits_{n=0}^\infty x^{n}+\sum\limits_{n=0}^\infty x^{n+1}=1+\sum\limits_{n=1}^\infty x^{n}+\sum\limits_{k=1}^\infty x^{k}=1+2\sum\limits_{n=1}^\infty x^{n}$$
A: There are two ways to look at this. First, as you have noted, 
$$
\begin{align}
\frac{1+x}{1-x}
&=\sum_{n=0}^\infty x^n+\sum_{n=0}^\infty x^{n+1}\\
&=1+\sum_{n=1}^\infty x^n+\sum_{n=1}^\infty x^n\\
&=1+2\sum_{n=1}^\infty x^n
\end{align}
$$
The second is to notice that
$$
\begin{align}
\frac{1+x}{1-x}
&=1+\frac{2x}{1-x}\\[6pt]
&=1+2x\sum_{n=0}^\infty x^n\\
&=1+2\sum_{n=1}^\infty x^n
\end{align}
$$
