Let $f(x)=\frac{1}{(1+x)^{1/2}}$. Supposed to find the Taylor series $Tf$ of $f$ around $0$, and show that it converges to $f$ on $[-1,1)$, (although I suspect there's a misprint in the book, and that this is supposed to be $(-1,1]$, since $f$ is not defined in $x=-1$).
If $f(x)=T_nf(x)+R_nf(x)$, where $T_nf(x)$ denotes the $n$th Taylor polynomial, then there's a result in my book which says, in short, that if the $(n+1)$-th derived $f^{(n+1)}$ of $f$ is such that there's a number $M$ with $|f^{(n+1)}(t)|\leq M$ for all $t\in(-1,1]$, then $|R_n f(x)|\leq \frac{M}{(n+1)!}|x|^{n+1}.$ My plan was to use this to show that $|R_n f(x)|\rightarrow 0$ as $n\rightarrow \infty$ for $x\in(-1,1]$. Can't seem to find any such $M$, however, and actually such an $M$ seems unlikely to me, seeing as, if I'm not mistaking, $|f^{(n+1)}(t)|=\frac{3*5*\cdots*(2n+1)}{2^{n+1}|1-t|^{\frac{2n+3}{2}}}.$ For doesn't this grow big when $t$ is close enough to $-1$?
I may however very well be on an entirely wrong track here...with the $R_n f$ argument and stuff.
Very thankful for any help.