Show that the Maclaurin series of $f(x)=1/\sqrt{1-x}$ holds for $x\in[0,1/2]$ using the Lagrange remainder theorem 
Show that the Maclaurin series of $f(x)=1/\sqrt{1-x}$ holds for $x\in[0,1/2]$ using the Lagrange remainder theorem

I can see that
$$f'(x)=\frac12 (1-x)^{-\frac32}\text{ and }f''(x)=\frac12\frac32(1-x)^{-\frac52}$$
and in general
$$f^{(n)}(x)=\frac{\prod_{j=1}^{n}\left(j-\frac12\right)}{(1-x)^{n+\frac12}}$$
Then the Maclaurin series can be written as the limit when $N\to\infty$ of
$$\mathcal M_N[f(x)]=\sum_{n=0}^{N}\frac{x^n\prod_{j=1}^{n}\left(j-\frac12\right)}{n!}=\sum_{n= 0}^{N}x^n\prod_{j=1}^{n}\left(1-\frac1{2j}\right)$$
And we can write the Lagrangian remainder as
$$\mathcal R_N[f(x)]=\frac{f^{(N+1)}(c_N)}{(N+1)!}x^{N+1}=\frac{x^{N+1}}{(1-c_N)^{N+\frac32}}\prod_{j=1}^{N+1}\left(1-\frac1{2j}\right)$$
for some $c$ such that $|c|<|x|$. Then my problem comes here: I can show, using the ratio test, that $\mathcal R_N[f(x)]$ converges for $x\in[0,1/2]$. But I think that this is not enough to show that the Maclaurin series holds on $[0,1/2]$ because the remainder must converges to zero, and I dont know how to show it.
My question: it is just enough to show that the remainder converges or I must show that it converges to zero? If it is the second case (as I think), how I can show it in this particular problem? Thank you in advance.

I had a mistake writing the formulas in the original question, now is corrected and is clear that the remainder converges to zero because the productory converges and $\left|\frac{x}{1-c_N}\right|<1$ because 
$$0\le |c_N|<|x|<1/2$$
Then now we have
$$\lim_{n\to\infty}\mathcal R_N[f(x)]=\lim_{n\to\infty}\left(\frac{x}{1-c_N}\right)^{N+1}(1-c_N)^{-\frac12}\prod_{j=1}^{N+1}\left(1-\frac1{2j}\right)\\<\lim_{n\to\infty}k\left(\frac{x}{1-c_N}\right)^{N+1}\left(\frac12\right)^{N+1}=0$$
with $k>0$.
