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

##### 1 Answer

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

I got

#sum_(n=0)^(N) (1+x)/(1-x)^2 = 1 + 3x + 5x^2 + 7x^3 + ...#

for some large finite

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

#1/(1-x) = sum_(n = 0)^N x^n = 1 + x + x^2 + x^3 + ...# (0)

for some large finite

#d/(dx)[-1/(1-x)] = 1/(1-x)^2# (1)

and that:

#color(darkred)(-1/(1-x)) = -sum_(n = 0)^N x^n# (2)

Analogizing from **(1)**:

#x/(1-x)^2 = x*["Power Series for "1/(1-x)^2]# (3)

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

First, notice how you can rewrite this as:

#(1+x)/(1-x)^2 = color(green)(1/(1-x)^2 + color(red)(x/(1-x)^2))# (4)

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

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

#color(green)(1/(1-x)^2) = d/(dx)[-1 - x - x^2 - x^3 - ...]#

#= color(green)(-1 - 2x - 3x^2 - ...)#

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

#color(red)(x/(1-x)^2) = x[-1 - 2x - 3x^2 - 4x^3 - ...]#

#= color(red)(-x - 2x^2 - 3x^3 - ...)#

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

# "Power Series of " (1+x)/(1-x)^2 = "Power Series of " color(green)(1/(1-x)^2 + color(red)(x/(1-x)^2))#

#prop [-1 - 2x - 3x^2 - 4x^3 - ...] + [-x - 2x^2 - 3x^3 - ...]#

#prop color(green)(-1 - 3x - 5x^2 - 7x^3 - ..).#

And finally, notice how we started with ** manipulated it** before taking the derivative to get

**The negative sign carried through all our operations.**

Therefore, to get the true power series, we have to ** undo** the multiplication we did on

#color(blue)(sum_(n=0)^(N) (1+x)/(1-x)^2 = 1 + 3x + 5x^2 + 7x^3 + ...)#

(that is why I purposefully put the