In order to calculate the Taylor-Maclaurin polynomial $\frac{1}{1+x}$ of order $n$ at point $0$, i used the identity:
$$\sum_{i=0}^n x^i + \frac{x^{n+1}}{1+x} = \frac{1}{1+x}\qquad (I)$$
and i admitted that the summation have $n+1$ elements while i want the Taylor-Maclaurin polynomial to be of order $n$.
Instead of take the derivative of order $k$ of the right member of the identity (I) i took the derivative of the left member, using the derivative properties i could calculate the derivative of $\sum_{i=0}^n x^i$ and i found to be $(-1)^{k}k!$ at point $0$, the trouble was to show that the $k$-order derivative of $\frac{x^{n+1}}{1+x}$ at $0$ is equal to $0$, i tried to use induction but i did not get anything. Does someone have any ideas ??
for $|x|<1$ : $$ \frac{1}{1-x}=\sum_{n\geq 0} x^n $$ put $t=-x$ you get : $$ f(t)=\frac{1}{1+t}=\sum_{n\geq 0} (-1)^n t^n=\sum_{n\geq 0} \frac{(-1)^nn!}{n!} t^n=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} t^n $$ so $$ f^{(n)}(0)=(-1)^n n! $$