How can I prove the general formula for Taylor Series? I am trying to prove the general form for practice and I am having some trouble with it. The question in my book is "Suppose f is infinitely differentiable on an open interval I containing the point x = c, and there exists a constant M > 0 such that |$f^{(n)} (x)$| $\leq$ M for all x $\in$ I and all n $\geq$ 0. Show that 
$$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!} (x-c)^n $$
My thoughts were to use the definition of a power series 
$$ \sum a_n (x-c)^n $$ 
and then to make the assumption that the power series converges with $|x-c| < M $ then to let $f(x)$ to be equal to our power series than evaluating each n-th derivative at $x=c$ than to make the statement that if we repeat this process infinitely many times we get the desired form of $a_n = \frac{f^{(n)}(c)}{n!} $ 
This is what I am unsure about because I am not sure if this would be strong enough to prove the Taylor's form especially with trying to get the form of $a_n$ any suggestions would be appreciated. Cheers
 A: Your approach is a reasonable one to gain intuition.  However, it tacitly assumes that one can differentiate the power series term by term as if it were a finite sum.  The fact that it is not a finite sum creates a need to apply a more rigorous way forward.  To that end we proceed.

Note that from Taylor's Theorem, we can write for a $n$ times differentiable function at $x=c$
$$f(x)=\sum_{k=0}^n\frac{f^{(k)}(c)}{k!}(x-c)^k+R_n(x,c) \tag1$$
where the remainder $R_n(x,c)=o(|x-c|^n)$ as $x\to c$. 
From $(1)$ we have $R_n(x,c)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(c)}{k!}(x-c)^k$,  
If for $x\in (c-a,a+c)$, $f$ and all of its derivatives exist, and if $x_0\in (c-a,c+a)$, then $f$ is represented by its Taylor series at $x_0$ if and only if
$$\lim_{n\to \infty}R_n(x,c)=0$$

There are a variety of explicit forms for the remainder term and estimates thereof.    The mean-value form of the remainder,
$$R_n(x,c)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-c)^{n+1}$$
for some $\xi\in(c,x)$.  If all derivatives of $f$ are bounded above with $|f^{(k)}|\le M$ for $|x-c|< a$, then
$$|R_n(x,c)|\le M\frac{a^{n+1}}{(n+1)!}$$
and the Taylor series converges uniformly to $f$.  In such as case, we can write
$$f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(c)}{k!}(x-c)^k$$
for $x\in (c-a,c+a)$
