Series representations: $$ \frac{1}{2(1-x)^3}+\frac{1}{4(1-x)^2}+\frac{1}{8(1-x)}+\frac{1}{8(1+x)}=\sum_{n=0}^\infty x^n\left(7+(-1)^n+8n+2n^2\right). $$
I'm trying to figure out to to turn this regular expression (to the left) to the generating function that WolframAlpha was so kind to provide it for me.
Or even just a way of finding the coefficient of a certain $x^n$ in the series...
By trying to simplify the expression I've got here : $$ -\frac{1}{(x-1)^3(x+1)}, $$ I can break this down to 2 known series: $$ \frac{1}{1+x} = (1-x+x^2-x^3+x^4-...) $$ multiplied by: $$ \frac{1}{(1-x)^3} = (1-x+x^2+x^3+x^4+...)^3. $$ But these are two series multiplied and I don't know how to continue from here to merge them both together to one series / generating function
I've been stuck on this for a couple of hours and I have no lead...
Thanks in advance
I think you are going in the opposite direction from the direction you want to be going. Although $\frac{1}{(1+x)(1-x)^3}$ is more compact, the expression in your first equation, which is the partial fraction decomposition of $\frac{1}{(1+x)(1-x)^3}$, is the more useful for getting the general term in the series.
The partial fraction decomposition is a sum of powers of binomials with constant coefficients. What you want to do now is to use the generalized binomial theorem to express the binomial powers as series. You already have the expansions of $(1-x)^{-1}$ and $(1+x)^{-1}$. You also need the expansions of $(1-x)^{-2}$ and $(1-x)^{-3}$. Combining the four terms with appropriate coefficients should give you the right hand side.