This question has already asked here Determine the set of all points where the Taylor series of the function $f(x)=\sum_{n=0}^{\infty} \frac {x^2}{(1+x^2)^n}$ around $x=e$ converges
I want to confirm my answer whether my answer is correct or not??
Q. Determine the set of all points where the Taylor series of the function $f(x)=\sum_{n=0}^{\infty} \frac {x^2}{(1+x^2)^n}$ around the point $x=e$ converges to $f(x)$.
My answer:
$f(x)=\sum_{n=0}^{\infty} \frac {x^2}{(1+x^2)^n}=x^2 \sum_{n=0}^{\infty} \frac {1}{(1+x^2)^n}=x^2 \frac {1}{1-\frac {1}{1+x^2}}=x^2 \frac {1+x^2}{x^2} \; \text {for} \; |\frac 1{1+x^2}| \lt 1 \; \text {i.e.} \; x \neq 0$ as the summation is geometric series.
$\therefore f(x)=1+x^2 \; \forall \; x \neq 0.$
Also $f(0)=0+0+0\cdots=0.$
Now $f(e)=1+e^2, f'(e)=2e,f''(e)=2$ and $f^{(n)}(e)=0 \; \forall \; n \ge 3.$
The Taylor series around $x=e$ of $f(x)$ is $(1+e^2)+\frac {(x-e)(2e)}{1!}+\frac {(x-e)^2(2)}{2!}.$
Thus the Taylor series is a polynomial on $x \neq 0$ and hence it converges to $f(x)$ on $x\neq 0$.
Is my solution is correct or not correct ??