# Help with manipulating power series.

1. Aug 16, 2015

### MidgetDwarf

So i am given (1+x)/(1-x)^2 and I have to put it into a power series. I know that 1/(1-x)= 1+x+x^2+x^3+...=∑x^n from 0 to infinity. I am having problems factoring series.

I differentiate 1/(1-x).

I get, 1/(1-x)^2= 1+2x+3x^2+...= ∑nX^(n-1) the sum from 1 to infinity.

rewriting this equation.

1/(1-x)^2=∑(n+1)X^n. The sum from 0 to infinity.

multiply both sides by (1+x)

I get (x+1)/(1-x)^2 = (1+x)∑(n+1)X^(n) , from 0 to infinity.

Then I distribute 1+x.

∑(n+1)X^(n) + ∑(n+1)X^(n+1) both sums have an index of 0.

My problem is that I have having trouble factoring using summation notation and I am not sure how my book factored these 2 sums into 1 to get the answer

∑(2n+1)X^n with the index of 0.

When I factored my previous work I got. ∑(n+1)(X^(n)[1+X]

which I cannot seem to to put into the form the book has it.

2. Aug 16, 2015

### FactChecker

Pick an exponent n and look at the terms with xn. From the first and second summation, respectively, they are (n+1)xn and nxn. Sum them to get the book answer.

3. Aug 16, 2015

### MidgetDwarf

hmm. Do you mean by finding a general formula for the partial sums? Ie. n=0 +n=1+n=3+....+n= whatever? from here look at a pattern and construct the formula?

What i am thinking: Is it kind of the way of how we play around with a telescopic series?

4. Aug 16, 2015

### micromass

Make the exponents of the $x$ equal and add.

5. Aug 16, 2015

### FactChecker

No. Nothing fancy. Just find the term in each summation that has the same xn and add them together. Combine like terms.