Given the function $$f(x)=\frac{3}{2-x-x^2}$$
- Expand the function into a McLaurin series
- Find a radius of convergence
- Find points, where the series convergent to the original function $f(x)$
What was done:
- We write $\frac{3}{2-x-x^2} = \frac{1}{1-x} + \frac{1}{2+x}$ and note that $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ where the series converges for $-1<x<1$. Note
$$\frac{1}{2+x} = \frac{1}{1-(-1-x)} = \sum_{n=0}^\infty (-1-x)^n = \sum_{n=0}^\infty (-1)^n(1+x)^n$$ where we can find radius by d'Alembert's rules: $$\lim_{n=\infty}\frac{(-1)^{n+1}(1+x)^{n+1}}{(-1)^n(1+x)^n}<1$$ meaning the series converges for $-1<x<0$. We write
$$\frac{3}{2-x-x^2}=\sum_{n=0}^\infty (-1)^n(1+x)^n + \sum_{n=0}^\infty x^n = \sum_{n=0}^\infty (x^n + (-1)^n(1+x)^n)$$
Suppose the fact that only sum of convergent series can be convergent, determine the radius as $-1<x<0$.
By how to find such points or interval? Is it possible that such an interval does not cover the entire radius of convergence?
Moreover I'm pretty sure that is something wrong in the existing part. I would appreciate any comments.