# How do you find a power series representation for  (1+x)/((1-x)^2)?

Sep 16, 2015

Fair warning---I expect this to be a long answer!

I got

${\sum}_{n = 0}^{N} \frac{1 + x}{1 - x} ^ 2 = 1 + 3 x + 5 {x}^{2} + 7 {x}^{3} + \ldots$

for some large finite $N$.

So, something that I believe you have already been taught is that:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{N} {x}^{n} = 1 + x + {x}^{2} + {x}^{3} + \ldots$ (0)

for some large finite $N$. Notice how:

$\frac{d}{\mathrm{dx}} \left[- \frac{1}{1 - x}\right] = \frac{1}{1 - x} ^ 2$ (1)

and that:

$\textcolor{\mathrm{da} r k red}{- \frac{1}{1 - x}} = - {\sum}_{n = 0}^{N} {x}^{n}$ (2)

Analogizing from (1):

$\frac{x}{1 - x} ^ 2 = x \cdot \left[\text{Power Series for } \frac{1}{1 - x} ^ 2\right]$ (3)

Remember these four relationships, because we will be referring back to them.

First, notice how you can rewrite this as:

$\frac{1 + x}{1 - x} ^ 2 = \textcolor{g r e e n}{\frac{1}{1 - x} ^ 2 + \textcolor{red}{\frac{x}{1 - x} ^ 2}}$ (4)

Already you may see how things could unfold. Making use of (0) in conjunction with (2), we get:

$\textcolor{\mathrm{da} r k red}{- \frac{1}{1 - x} = - 1 - x - {x}^{2} - {x}^{3} - \ldots}$

Now, making use of (1), we get:

$\textcolor{g r e e n}{\frac{1}{1 - x} ^ 2} = \frac{d}{\mathrm{dx}} \left[- 1 - x - {x}^{2} - {x}^{3} - \ldots\right]$

$= \textcolor{g r e e n}{- 1 - 2 x - 3 {x}^{2} - \ldots}$

This becomes the left half that we are looking for. Now for the right half. Using (3), we get:

$\textcolor{red}{\frac{x}{1 - x} ^ 2} = x \left[- 1 - 2 x - 3 {x}^{2} - 4 {x}^{3} - \ldots\right]$

$= \textcolor{red}{- x - 2 {x}^{2} - 3 {x}^{3} - \ldots}$

Next, we can add them together (like-terms with like-terms) according to (4):

$\text{Power Series of " (1+x)/(1-x)^2 = "Power Series of } \textcolor{g r e e n}{\frac{1}{1 - x} ^ 2 + \textcolor{red}{\frac{x}{1 - x} ^ 2}}$

$\propto \left[- 1 - 2 x - 3 {x}^{2} - 4 {x}^{3} - \ldots\right] + \left[- x - 2 {x}^{2} - 3 {x}^{3} - \ldots\right]$

$\propto \textcolor{g r e e n}{- 1 - 3 x - 5 {x}^{2} - 7 {x}^{3} - . .} .$

And finally, notice how we started with $\frac{1}{1 - x}$, but we manipulated it before taking the derivative to get $\frac{1}{1 - x} ^ 2$ by multiplying it by $- 1$ to get the power series for $- \frac{1}{1 - x}$.

The negative sign carried through all our operations.

Therefore, to get the true power series, we have to undo the multiplication we did on $\frac{1}{1 - x}$ by dividing by $- 1$ at the end, thus dividing the previous result by $- 1$. So, finally, we get:

$\textcolor{b l u e}{{\sum}_{n = 0}^{N} \frac{1 + x}{1 - x} ^ 2 = 1 + 3 x + 5 {x}^{2} + 7 {x}^{3} + \ldots}$

(that is why I purposefully put the $\propto$ symbol to emphasize that we were not at the final answer yet)

Wolfram Alpha agrees with this answer.