Taylor Series of $f(x) =\frac{1}{x^2}$

Question

I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.

$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$

I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$

So I got the expansion:

$$f(x)=1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+...$$

What is difficult for me is expressing this in summation notation. Is this right?

$$\sum\limits_{n=0}^\infty (n+1)(x+1)^n$$

Answer 1

You can verify it by computing this sum, because for $|x+1|<1$ this infinite sum is convergent: $$ \frac{d}{dx}(x+1)\sum_{k=0}^{\infty} (x+1)^{k+1} = \frac{d}{dx}\bigg(-\frac{x+1}{x} - 1 \bigg) = \frac{d}{dx} \bigg( -\frac{1}{x} \bigg) = \frac{1}{x^2} $$

Answer 2

For $x$ such that $|x+1|<1$

$$f(x)=\frac{-1}{x}=\frac{1}{1-(x+1)}$$

$$=1+(x+1)+(x+1)^2+....$$

$$=\sum_{k=0}^\infty (x+1)^k$$

As a sum of a power series, $f $ is differentiable at $ (-2,0)$ and

$$f'(x)=\frac{1}{x^2}=\sum_{k=1}^\infty k(x+1)^{k-1}$$

$$=\sum_{k=0}^\infty(k+1)(x+1)^k$$