<s>[INST] Taylor series for 1/(1-x). I am trying to write a Taylor series for $$ f(x)=\frac{1}{1-x}, \ x<1  \ .$$
In most sources, it is said, that this function can be written as a Taylor series, if $$ \left| x \right|<1. $$
However, I don't get the same condition for x.
Because $$ f^{\left(n\right)}\left(x\right)=\frac{n!}{\left(1-x\right)^{n+1}},$$ the remainder term (or its abolute value) is $$ \left|\frac{x^{n+1}}{\left(1-c\right)^{n+2}}\right|=\frac{\left|x\right|^{n+1}}{\left|1-c\right|^{n+2}}{,}\ \mathrm{where} \ 0\le c\le x<1\ \mathrm{or}\ x\le c\le0.$$
If $$ 0 \le c \le x<1, $$ then
$$  0\ge-c\ge-x>-1\ \Leftrightarrow\ 1\ge1-c\ge1-x>0\ \Leftrightarrow\ 0<\left(1-x\right)^{n+2}\le\left(1-c\right)^{n+2}\le1\ \Leftrightarrow\ 1\le\frac{1}{\left(1-c\right)^{n+2}}\le\frac{1}{\left(1-x\right)^{n+2}} \Leftrightarrow\ \left| x^{n+1} \right|\le \left| \frac{x^{n+1}}{\left(1-c\right)^{n+2}} \right| \le \left| \frac{x^{n+1}}{\left(1-x\right)^{n+2}} \right|=\begin{cases}
e^{\ln\frac{x}{1-x}\cdot n+\left(\ln x-2\ln\left(1-x\right)\right)}\ , \ 0<x<1&\\
0\ , \ x=0&
\end{cases} .$$
When  $$ n \rightarrow \infty, $$ for remainder to converge to zero, x has to be such that $$ 0 \le x < \frac{1}{2}.$$
Then the lower bound of the remainder $$   \left|x^{n+1}  \right| \rightarrow 0 $$ and the upper bound $$   \left| \frac{x^{n+1}}{\left(1-x\right)^{n+2}}\right|  \rightarrow 0 .$$
If $$ x \le c \le 0 ,$$ then
$$ \ -x\ge-c\ge0\ \Leftrightarrow\ 1-x\ge1-c\ge1\ \Leftrightarrow\ \left(1-x\right)^{n+2}\ge\left(1-c\right)^{n+2}\ge1\ \Leftrightarrow\ \frac{1}{\left(1-x\right)^{n+2}}\le\frac{1}{\left(1-c\right)^{n+2}}\le1\ \Leftrightarrow\ \left|x\right|^{n+1}\ge\left|\frac{x^{n+1}}{\left(1-c\right)^{n+2}}\right|\ge\left|\frac{x^{n+1}}{\left(1-x\right)^{n+2}}\right|=\frac{\left|x\right|^{n+1}}{\left|1-x\right|^{n+2}}=\begin{cases}
e^{\ln\left|\frac{x}{1-x}\right|\cdot n+\left(\ln\left|x\right|-2\ln\left|1-x\right|\right){,}\, x<0}&\\
0\ , \ x=0&
\end{cases} . $$
When  $$ n \rightarrow \infty, $$ for remainder to converge to zero, x has to be such that $$ -1 < x \le 0.$$
Then the upper bound of the remainder $$\left|x\right|^{n+1} \rightarrow 0 $$ and the lower bound $$  \frac{\left|x\right|^{n+1}}{\left|1-x\right|^{n+2}} \rightarrow 0. $$
Thus I would say, that function f can be written as Taylor series only when $$  -1 < x < \frac{1}{2} . $$ Isn't this true?
 [/INST] It is a well-known and indisputable (even elementary) fact that for all $x\ne1$ and all natural $n$
$$\sum_{k=0}^n x^k=\frac1{1-x}-\frac{x{^{n+1}}}{1-x},$$
implying that $\forall|x|<1$,
$$\lim_{n\to\infty}\sum_{k=0}^n x^k=\frac1{1-x}.$$
This falsifies your claim.

Needless to say,
$$\left.\left(\frac1{1-x}\right)^{(k)}\right|_{x=0}=k!$$

Note that the remainder can be evaluated exactly as
$$\sum_{k=n+1}^\infty x^k=\frac{x^{n+1}}{1-x}.$$
Comparing to the Taylor-Lagrange expression of the remainder,
$$\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\frac{x^{n+1}}{(1-c)^{n+2}},$$
we have
$$c=1-\sqrt[n-2]{1-x}\le1.$$
This formula shows that a partial sum plus the remainder yield a correct evaluation  of the function for all $x<1$ and cannot work for $x\ge1$ (because of the singularity).
In fact, the study of the Taylor-Lagrange remainder cannot tell you about convergence.
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