Given this equivalence: $$\frac{1}{z^3+z+2}=\sum_{n=-\infty}^\infty (z+1)^n c_n$$ I must determine radius of convergence of the series and the coefficients $c_n$.
I'm definitely unsure of what to do here. I think that the radius of convergence is $$R=\frac{1}{\limsup_{n\rightarrow\infty}{\sqrt[n]{|c_n|}}}$$ Which of course I can't compute without having the coefficients. So I've looked at the sum of the series and evaluated it's roots, posing $t=z+1$: $$\frac{1}{(z+1)(z-\frac{1}{2}-\frac{i}{2}\sqrt7)(z-\frac{1}{2}+\frac{i}{2}\sqrt7)}=\frac{1}{t(t-\frac{3}{2}-\frac{i}{2}\sqrt7)(t-\frac{3}{2}+\frac{i}{2}\sqrt7)}$$ And I think that it should be $|t|<2$, $|z+1|<2$.
I was looking for a recurrence law but I'm unsure and I haven't found out a lot but I'll show some of my tries. $$1=(z^3+z+2)\sum_{n=-\infty}^\infty (z+1)^n c_n=(t^3-3t^2+4)\sum_{n=-\infty}^\infty t^n c_n=(?)= \sum_{n=-\infty}^\infty t^n (c_{n-3}-3c_{n-2}+4c_n)$$ I'm perplexed by that $-\infty$ from wich the series starts..
Set $w=z+1$, then $$ z^3+z+2=w^3-3w^2+3w+w=w(w^2-3w+4) $$ so that there are two poles on the circle $|w|=2$, which gives you the radius of convergence for the power series in $w=z+1$.
Using a partial fraction decomposition on the factorization in linear factors and then the geometric series allows to find explicit formulas for the coefficients.
You get the $-\infty$ as lower index bound as a sign that the solution might not be a power series but a more general Laurent series. And indeed, as $w$ is a factor, the first term in the series has power $-1$. $$ w\sum_{n=-1}^\infty c_nw^n=\frac1{w^2-3w+4}=\frac{i}{\sqrt7}\left(\frac{1}{w-\frac32+i\frac{\sqrt7}2}-\frac{1}{w-\frac32-i\frac{\sqrt7}2}\right) $$