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Let $f:I \rightarrow \mathbb{R}$ be a function that can be diferentiated infinite times. Show that if exists $K>0$ s.t. $|f^{(n)}(x)|\leq K\; \forall\; x \in I$ and $n\geq 0$, then

$$f(x) = \sum^{\infty}_{i=0}\dfrac{f^{(i)}(x_0)}{i!}(x_0-x)^i$$ $\forall x, x_0 \in I$

I must show that

$$S_n = \sum^{n}_{i=0}\dfrac{f^{(i)}(x_0)}{i!}(x_0-x)^i$$ converges to $f(x)$.

In other words, let $\epsilon>0$, show that exists $N>0$ s.t. $|S_n-f(x)|\leq \epsilon\; \forall \;n\geq N$.

My idea was to write $f(x)$ as a taylor polynomial of order $N$ plus a remainder and somehow show that this remainder would be less than $\epsilon$. That is $f(x) = S_N+r(x-x_0)$ where

$$r(x-x_0) = \sum^{\infty}_{i=N+1}\dfrac{f^{(i)}(x_0)}{i!}(x_0-x)^i$$

Now, how do I find the $N$ and how to show that $r(x-x_0)\leq \epsilon$?

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    If $|x - x_0| < c$ then $r(x - x_0) \leq \sum_{i > N} \dfrac{K c^i}{i!} = t_N;$ if $a_i = \dfrac{c^i}{i!},$ then $(a_i)$ defines a convergent _series_ and so, it's tail, namely $t_N,$ converges to zero. Is this of help to you? – 2017-01-03
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    Yes it helps! Thank you @WillM. – 2017-01-03

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