So you are trying to prove that your function is equal to its power series representation, and as you said to do this you will use Taylors theorem and inequality.
$\mathbf{Thereom:}$ If $f(x)=T_n(x)+R_n(x) , $ where $T_n$ is the $n^{th}$ degree taylor polynomial of $f$ at $a$ and $\lim_{n\to \infty} R_n(x)=0$ for $|x-a| \lt R$ ,
then $f$ is equal to the sum of its Taylor series on the interval $|x-a| \lt R$ ( theorem as given in Stewart 7ed.)
And we are wanting to take advantage of the following
$\mathbf{Lemma}:$ If $| f^{n+1}(x)| \le M$ for $|x-a| \le d$ , then the remainder $R_n(x)$ of the Taylor series satisfies the inequality
$$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$$ for |$x-a| \le d$
So what is the general form of the taylor series for $f(x)=\frac{1}{x}$ centred at $x=1$?
As you said , it has the general form $\sum_{n=0}^{\infty}(-1)^{n}(x-1)^{n}$ and converges for $|1-x| \lt 1$
Now , $f(x)=(1/x)$ and we have that $f^{n}(1)=(-1)^{n}n!$
$f^{n+1}(1)=(-1)^{n+1}(n+1)! \le (n+1)! $
$|x-a| \le 1$
and by the lemma we have $| R_n(x)| \le \frac{(n+1)!}{(n+1)!}|x-a|^{n+1}$
Do you see anything we could do from here?
So apparently the reason is fairly simple:
The given Taylor Polynomial is not a series, so its interval of convergence is just the center—2 in this case.
Therefore, without further proof/explanation, one could not use $T’$ to determine if $f’$ is positive or negative at any point other than the center ($x=2$).
Best Answer
What you're looking for is a non-analytic smooth function. See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function