I am having trouble determining the Taylor series and convergence interval for $f(x)=1/x$ around $x=1$. I have attempted the problem, but I'm not sure if my solution is correct.
Here's what I did: I first found the derivatives of $f(x)=1/x$: $$f(x)=1/x$$ $$f'(x)=-1/x^2$$ $$f''(x)=2/x^3$$ $$f'''(x)=-6/x^4$$ $$f^{(4)}(x)=24/x^5$$ $$\vdots$$ $$f^{(n)}(x)=(-1)^{n+1}n!/x^{n+1}$$
Then I used the formula for the Taylor series of $f(x)$ around $x=1$: $$\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n$$
Substituting the derivatives into the formula, I get: $$\sum_{n=0}^\infty \frac{(-1)^{n+1}}{n!}(x-1)^n$$
For the convergence interval, I used the ratio test: $$\lim_{n\to\infty} \left|\frac{(-1)^{n+2}}{(n+1)!}\frac{n!}{(-1)^{n+1}}\frac{(x-1)^{n+1}}{(x-1)^n}\right|=\lim_{n\to\infty}\frac{|x-1|}{n+1}=0$$
Therefore, the series converges for all $x$.
Now I'm not sure if my solution is correct or if there's any mistake. Any help would be greatly appreciated.
Thank you in advance!