Representing Functions as Power Series Rewrite $$f(x)=(1+x)/(1-x)^2$$ as a power series.
Work thus far:
I separated it into two parts:
$$1/(1-x)^2 + x/(1-x)^2$$
I realize that the first expression is the derivative of $1/(1-x)$ and come up with this sum of series:
$$\sum_{n=0}^\infty x^n$$
Since a derivative was involved, we must derive the series:
$$\displaystyle\sum_{n=0}^\infty nx^{n-1} + x \displaystyle\sum_{n=0}^\infty nx^{n-1}$$
$$=\displaystyle\sum_{n=0}^\infty nx^{n-1} + \displaystyle\sum_{n=0}^\infty nx^n$$
These series need to be added, but how? 
The final answer is $\displaystyle\sum_{n=0}^\infty (2n+1)x^n$
 A: Shift the indices so that the powers of $n$ inside match:
$$\begin{align*}
\sum_{n\ge 0}nx^{n-1}+\sum_{n\ge 0}nx^n&=\sum_{n\ge 1}nx^{n-1}+\sum_{n\ge 0}nx^n\\\\
&=\sum_{n\ge 0}(n+1)x^n+\sum_{n\ge 0}nx^n\\\\
&=\sum_{n\ge 0}(2n+1)x^n\;.
\end{align*}$$
It may be easier to understand at first if you write out a few terms:
$$\left(1+2x+3x^2+\ldots+(n+1)x^n+\ldots\right)+\left(x+2x^2+3x^3+\ldots+nx^n+\ldots\right)$$
Here you can see that the coefficient of $x^n$ in the first sum is $n+1$, and the coefficient of $x^n$ in the second sum is $n$, so when you combine the sums the coefficient of $x^n$ must be $2n+1$.
If you find the shift
$$\sum_{n\ge 1}nx^{n-1}=\sum_{n\ge 0}(n+1)x^n$$ 
a bit mysterious, do it with an intermediate step: let $k=n-1$, so that $n=k+1$, and substitute to get
$$\sum_{n\ge 1}nx^{n-1}=\sum_{k+1\ge 1}(k+1)x^k=\sum_{k\ge 0}(k+1)x^k\;,$$
and then just rename the index variable back to $n$.
A: Expand $(1-x)^2$ into $(1-2x+x^2)$. 
Then use long division to divide $(1-2x+x^2)$ into $(1+x)$.
Stop when you have enough terms in the quotient to identify the pattern.
Seems too easy??
A: $$
\frac{1}{1-x}=1+x+x^2+x^3+\ldots
$$
$$
\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\ldots
$$
$$
\frac{1+x}{(1-x)^2}=1+2x+3x^2+\ldots +x+2x^2+ 3x^3+\ldots=1+3x+5x^2+7x^3+\ldots
$$
