# Proof with Lagrange's Remainder Theorem

Consider the function $f(x) = \frac{1}{\sqrt{1-x}}$. Generate the Taylor series for f centered at zero, and use Lagrange's Remainder Theorem to show the series converges to f on $[0,1/2].$

I have generated the Taylor series, which is $$1+\frac{x}{2} + \frac{3x^2}{8}+\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}+...$$ I'm still pretty shaky on using Lagrange's Remainder Theorem to prove convergence, and it doesn't seem very straightforward with this problem. For reference, LRT says that given a point $x$ in the radius of convergence of the series, there exists a $c$ satisfying $|c|<|x|$ where the error function $E_N(x)$ satisfies $$E_N(x)=\frac{f^{N+1}(c)}{(N+1)!}x^{N+1}$$

We have $f(x) = (1-x)^{-\alpha}$ where $\alpha = 1/2$.

Note that

$$|E_{N}| = \left| \frac{\alpha(\alpha-1) \ldots (\alpha - N)x^{N+1}}{(N+1)!}(1-c)^{-\alpha - N -1}\right|\\ = \alpha|x||1-c|^{-\alpha-1}\left| \frac{(\alpha-1) \ldots (\alpha - N)}{(N+1)!}\left(\frac{x}{1-c}\right)^N\right|.$$

Consider

$$a_Nz^N = \frac{(\alpha-1) \ldots (\alpha - N)}{(N+1)!}\left(\frac{x}{1-c}\right)^N.$$

Since $0 < c < x < 1/2$ we have $|z| < 1$ and

$$\lim_{N\to \infty} \frac{|a_{N+1}z^{N+1}|}{|a_N z^N|} = \lim_{N\to \infty} \frac{|\alpha - N - 1|}{N+2}|z| = |z| < 1.$$

By the ratio test $\sum a_Nz^N$ converges and

$$\lim_{N \to \infty}|a_N z^N| = 0.$$

Hence, $|E_{N}| \to 0$ and the Taylor series for $f$ converges.