$f(x)= \sum_\limits{k=0}^{n} \dfrac{f^k(a)}{k!}(x-a)^k + \dfrac{ f^{n+1}(c) }{ (n+1)! } (x-a)^{n+1} $
I have some trouble understanding this, as it seems to imply that
$ \dfrac{ f^{n+1}(c) }{ (n+1)! } (x-a)^{n+1} = \sum_\limits{k=n+1}^{\infty} \dfrac{f^k(a)}{k!}(x-a)^k$
Because by the Taylor expansion of $f(x)$ at a:
$f(x)= \sum_\limits{k=0}^{ \infty} \dfrac{f^k(a)}{k!}(x-a)^k $
This seems to be very similar to the mean value theorem, but I'm not sure how to prove the equation using it as it includes x to powers other than 1, and I only know $\dfrac{f(b)-f(a)}{b-a}=f'(a \le k \le b) $. As it may be of help here for me to understand the answer: how can the mean value theorem be derived from the Taylor series (if it can at all), and can this be generalised to higher derivitaves (how?)?