I am trying to manually present the function $f(x) = \sqrt{1-x}\ \ $(for $0 \le x \lt 1$) as a power series and to prove that the power series converges to the function value $f(x)$.
I am using Maclaurin's power series formula.
For the n-th derivative ($n \ge 2$) I got this
$$f^{(n)}(x) = \frac{(-1).1.3.5 \dots (2n-3)}{2^n} . (1-x)^{-(2n-1)/2}$$
I hope it's correct. Is it?
Then using that I was able to compute the power series and to prove that it converges.
But in order to prove that it converges to the function value $f(x)$, I think I need to prove that the remainder term converges to zero as $n \to \infty$
So I have these questions. Is my remainder term (which I computed by hand) correct?
$$R_n = \frac{(-1)1.3.5 \dots (2n-1)}{2^{n+1}(n+1)!} . \frac{x^{n+1}}{(1-\theta x)^{(2n+1)/2}}$$
Here we just know that $\theta$ is some number, $\theta \in (0,1)$ and $\theta$ depends on $x$.
And then how do I prove that $R_n$ goes to zero as $n \to \infty$ ?
Since $\sqrt~$ is analytic on $\Bbb C\setminus(-\infty,0],$ you get for free that the limit of the Taylor series of $f$ (on its disc of convergence) is $f.$ You don't have to bother about $R_n.$ Just compute the coefficients of the Taylor series of $f$: $$\sqrt{1-x}=1-\sum_{k=0}^\infty\frac2{k+1}\binom{2k}{k}\left(\frac x4\right)^{k+1}$$ and derive its radius of convergence, which is equal to $1.$