Taylor polynomial - error doesn't converge always to $0$

I want to calculate the Taylor polynomial of order $n$ for the funktion $f(x) = \frac{1}{ 1−x}$ for $x_0=0$ and $0 < x < 1$ and the remainder $R_n$.

I have found that \begin{equation*}P_{0,n}(x)=\sum_{k=0}^n\frac{f^{(k)}(0)\cdot x^k}{k!}=\sum_{k=0}^n x^k \end{equation*}

Is the remainder \begin{equation*}R_n=\frac{\frac{(n+1)!}{(1-\xi)^{n+2}}}{(n+1)!}\cdot (x-0)^n=\frac{x^n}{(1-\xi)^{n+2}}\end{equation*} ? Or do we have to use the fomrula with the integral?

In some notes I read that in general the remainder doesn't converge to $0$ for all $\xi\in (0,x)$. Can you give me an example for that?

• The remainder of " basic " function like logarith, exponential does converge to $0$. – Atmos Jan 21 '18 at 20:26
• Ah ok! When it is asked to calculate the remainder do I have to use the formula I used in the initial post? Becayuse I have seen that there is also a formula with integral. @Atmos – Mary Star Jan 21 '18 at 20:29
• When we consider the function $g(x)=e^x$, we get the remainder \begin{equation*}R_n=\frac{g^{(n+1)}(\xi )}{(n+1)!}\cdot x^n=\frac{e^{\xi}}{(n+1)!}\cdot x^n\end{equation*} right? We have to calculate the limit as $n\rightarrow \infty$, right? @Atmos – Mary Star Jan 21 '18 at 20:32
• Was speaking about integrals form of the remainder – Atmos Jan 21 '18 at 20:52
• Ah! So, does this only hold for the integral form, or are there other examples where the remainder in the form I used does not converge to $0$ ? @Atmos – Mary Star Jan 21 '18 at 20:53

The Lagrange form of the remainder is $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} = \frac{x^{n+1}}{(n+1)(1-\xi)^{n+1}}$
For this particular $f$, the remainder is easy to compute exactly as $$f(x)-\sum_{k=0}^n x^k=\frac{1}{1-x}-\sum_{k=0}^n x^k=\frac{x^{n+1}}{1-x}.$$
Therefore $R_n(x)=\frac{x^{n+1}}{1-x}$.