Power series representation of $\frac{1+x}{1-x}$ explanation? I don't really get how any of this works. I tried looking at this power function example already to figure it out. But I got lost when this part comes up.  $$\sum_{n=0}^{\infty}x^n=x^0+\sum_{n=1}^{\infty}x^n=1+\sum_{n=1}^{\infty}x^n$$ 
Why is this entire step happening and why is the $n$ index moving? And do you just distribute the $\sum_{n=1}^{\infty}x^n$ into $(1+x)$? 
 A: You have a sum of two geometric series.
\begin{align}
\frac 1 {1-x} & = 1 + x + x^2 + x^3 + \cdots \\[10pt]
\frac x {1-x} & = x + x^2 + x^3 + x^4 + \cdots
\end{align}
You're looking for the sum of the two above.
\begin{align}
\sum_{n=0}^\infty x^n & = x^0 + x^1 + x^2 + x^3 + x^4 + \cdots \\[10pt]
& = x^0 + \Big( x^1 + x^2 + x^3 + x^4 + \cdots \Big) \\[10pt]
& = x_0 + \sum_{n=1}^\infty x^n \\[10pt]
& = 1+\sum_{n=1}^{\infty}x^n
\end{align}
A: Sigma summation is simply a collection of terms that are added together, so the step you are looking at is basically $(x^0+x^1+x^2+\cdots+x^n)=x^0+(x^1+x^2+\cdots+x^n)=1+(x^1+x^2+\cdots+x^n)$
A: As you know:
$$\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$$
You are trying to find the power series representation of $\frac{1+x}{1-x}$.
First thing to do is to represent the fraction in a form similar seen at the beginning:
$$\frac{1+x}{1-x} = \frac{1}{1-x} + \frac{x}{1-x} = \sum_{n=0}^{\infty}x^n + \frac{x}{1-x}$$
Now we have to deal with the term $\frac{x}{1-x}$:
If we divide by $x$ on the top and bottom we get:
$$\frac{x}{1-x} = \frac{1}{\frac{1}{x} - 1} = -\frac{1}{1 -\frac{1}{x}} = -\sum_{n=0}^{\infty}\frac{1}{x^n}$$
Now we can represent our sum as:
$$\frac{1+x}{1-x} = \sum_{n=0}^{\infty}\left(x^n - \frac{1}{x^n}\right)$$
