Let $f(x) = \frac{1}{x}$ on $(0,2)$. I know the Taylor expansion of $f(x)$ at $1$ is: $p_n(x)=\sum_{k=0}^{\infty}(-1)^k(x-1)^k$. And I want to show that the reminder is : $f(x)-p_n(x)$= $\frac{(1-x)^{n+1}}{x}$ for $x\in (0,2)$.
But I cannot proceed by the Taylor reminder theorem. Do I need to apply the Abel theorem for power series?
Since $P_{n+1}(x)= \sum_{k=0}^{n+1} (-1)^k (x-1)^k$ we have $$R_{n+1}=\sum_{k=n+2}^{\infty} (-1)^k (x-1)^k= \frac{1}{(-1)^{n+2}(x-1)^{n+2}}\sum_{k=0}^{\infty} (-1)^k (x-1)^k\stackrel{\text{g. S.}}{=}\frac{1}{(-1)^{n+2}(x-1)^{n+2}} \frac{1}{1-x}.$$ In the last step we use that $\sum_{k=0}^{\infty} (-1)^k (x-1)^k= \sum_{k=0}^{\infty} (1-x)^k$ is a geometric series for $x\in (0,2)$.