Power Series Representation of $(1+x)/(1-x)$
Olivia Zamora 
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 $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1}$
$\endgroup$ 13 Answers
$\begingroup$$$f(x)=\frac{1+x}{1-x}=(1+x)\frac{1}{1-x}$$ $$=(1+x)\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}x^n+x\sum_{n=0}^{\infty}x^n=$$ $$=\sum_{n=0}^{\infty}x^n+\sum_{n=0}^{\infty}x^{n+1}$$ 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 $$(1+x)\frac{1}{1-x}=1+\sum_{n=1}^{\infty}x^n+\sum_{n=1}^{\infty}x^n=$$ $$=1+2\sum_{n=1}^{\infty}x^n$$
$\endgroup$ 3 $\begingroup$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 $$ \begin{align} \frac{1+x}{1-x} &=1+\frac{2x}{1-x}\\[6pt] &=1+2x\sum_{n=0}^\infty x^n\\ &=1+2\sum_{n=1}^\infty x^n \end{align} $$
$\endgroup$ 1 $\begingroup$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 $$\displaystyle\sum\limits_{n=0}^\infty x^{n}+\sum\limits_{n=0}^\infty x^{n+1}=1+\sum\limits_{n=1}^\infty x^{n}+\sum\limits_{k=1}^\infty x^{k}=1+2\sum\limits_{n=1}^\infty x^{n}$$
$\endgroup$ 4
