Trouble with form of remainder of $\frac{1}{1+x}$ While asking a question here I got the following equality
$\displaystyle\frac{1}{1+t}=\sum\limits_{k=0}^{n}(-1)^k t^{k}+\frac{(-1)^{n+1}t^{n+1}}{1+t}$
I'm trying to prove this with Taylor's theorem, I got that $f^{(n)}(x)=\displaystyle\frac{(-1)^nn!}{(1+x)^{n+1}}$ so the first part of the summation is done but I can't seem to reach the remainder according to Rudin the form of the remainder is  $\displaystyle\frac{(-1)^n!}{(1+c)^{n+1}}t^{n+1}$ . I know the equality can be proved by expanding the sum on the right but I want to get the result with the theorem. 
 A: One cannot obtain the desired equality  from Taylor's Theorem, except in a trivial sense. That result is a generalized Mean Value Theorem. Because of the presence of the "$c$", by direct use of Taylor's Theorem we can at best get estimates. Typically, these estimates are not even very good.
Remarks: $1.$ You are undoubtedly familiar with the procedure for finding $1+u+\cdots +u^n$. Call this sum $S$. Then $Su=u+u^2+\cdots +u^{n+1}$. Subtract. We get 
$$(1-u)S=1-u^{n+1}.$$
This gives $S=\dfrac{1-u^{n+1}}{1-u}=\dfrac{1}{1-u}-\dfrac{u^{n+1}}{1-u}$, which is a variant of your expression (put $u=-t$).
$2.$ By multiplying the desired equality through by $1+t$, we obtain a simple polynomial identity. A polynomial identity can always be proved trivially using Taylor's Theorem, by verifying that an appropriate higher order derivative is constant. 
A: Taylor will not help you here.
Hint: $\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}$ for all positive integers $n$, when $r\neq 1$.
Take $r=$...
A: As I answered your other question by saying the Taylor theorem is the way, I feel inclined to answer this one:
With the integral form of $R_n$:
$$R_n=\frac{1}{n!}\int_{0}^{x}f^{(n+1)}(t)(x-t)^n\, dt=(-1)^{n+1}(n+1)\int_{0}^{x}\frac{(x-t)^n}{(1+t)^{n+2}}dt$$
You must show
$$\int_{0}^{x}\frac{(x-t)^n}{(1+t)^{n+2}}dt=I=\frac{x^{n+1}}{(1+x)(n+1)}$$
With the substitution $u=1+t$ we have
$$I=\int_{1}^{x+1}\frac{(x+1-u)^n}{u^{n+2}}du=\int_{1}^{x+1}\left(\frac{x+1}{u}-1\right)^n\frac1{u^{2}}du$$
Substituting $v=\frac{x+1}u$,
$$I=-\int_{x+1}^{1}\left(v-1\right)^ndv=\int_{1}^{x+1}\left(v-1\right)^ndv$$
Things should be straightforward from here
