using taylor's theorem

Question

using Taylor's theorem . I need to find $f\left ( x \right ) = \frac{x^{2}+1}{x^{2}-2x+1}$ when $a=0$ and (the variable): $k=n$

I tried to use the taylor's theorem on $g\left ( x \right ) = \frac{1}{(x-1)^{2}}$

and i came to this result $f\left ( x \right ) =2\sum_{n=0}^{\infty}\left (n+1 \right )x^{n}+\sum_{n=0}^{\infty}\left (n+1 \right )x^{2}+\epsilon (x^{n})$ is it correct ?

thanks for your support.

Answer 1

More quickly note that

$$g(x)=\frac d{dx}\frac1{1-x}=\frac d{dx}\sum_{n=0}^\infty x^n=\sum_{n=0}^\infty(n+1)x^n$$

Multiply it by $x^2+1$ and you should've got

$$f(x)=\sum_{n=0}^\infty(n+1)x^{n+2}+\sum_{k=0}^\infty(k+1)x^k\\=1+2x+\sum_{n=2}^\infty(n-1)x^n+\sum_{k=2}^\infty(k+1)x^k\\=1+2x+2\sum_{n=2}^\infty nx^n$$

$$f(x)=1+2\sum_{n=1}^\infty nx^n$$

Answer 2

Hint: $$f\left ( x \right ) = \frac{x^{2}+1}{x^{2}-2x+1}=1+\frac{2}{(x-1)}+\frac{2}{(x-1)^2}$$