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.
Best Answer
If the remainder term $R_N$ does not converge to $0$ then the power series will not sum to $f(x).$
For just the case $0\leq x<1/2 :\quad$ Let $x=1/2-y.$ Then $c_N$ (which you write as $c$) satisfies $1-c_N>1-x$ for every $N. $ Let $z=x/(1-x).$ Then $0\leq z<1. $
For every $N$ we have $$0<x/(1-c_N)<z<1.$$ Note that $z$ does not depend on $N.$
Now since $0\leq z<1, $ there exists $k$ with $0<k<1$ such that for all sufficiently large $N$ we have $z(1+3/(2 N+2))<k.$ So for all sufficiently large $N$ we have $0<R_{N+1}<k R_N,$ so $R_N\to 0.$
That is, choose $N_0$ such that $n> N_0\implies z(1+3/(2 n+2))<k. $ Then for all $m> 0$ we have $0<R_{N_0+m}<k^m R_{N_0}.$