Suppose $f$ and all its derivatives exist on an open interval $I$ containing $a$. If there exists $M>0$ such that for all $x\in I, n\in\mathbb N$, $|f^{(n)}(x)|\le M^n$, prove $\lim_{n\to\infty}T_n(x)=f(x)$.
Here I am trying to prove that the Taylor Polynomial converges to $f(x)$ given that $f$ and all its derivatives are bounded. I unfortunately am doubtful of one of the steps in my proof:
PROOF: Observe $$\lim_{n\to\infty}\left|{{|x-a|^{n+1}\over(n+1)!}\over{|x-a|^n\over n!}}\right| = \lim_{n\to\infty}{|x-a|\over(n+1)} = 0 < 1.$$ Hence ${|x-a|^{n+1}\over(n+1)!}\to0$ as $n\to\infty$. In other words, for any $\epsilon > 0$ there exists $N\in\mathbb N$ such that for all $x\in I$ and for all $n\ge N$, we have $${|x-a|^{n+1}\over(n+1)!} < {\epsilon\over M^{n+1}} \;\;\;(!!!)$$
Given that $T_n(x)$ is the $n^{th}$ Taylor polynomial, then $$\begin{align}|f(x) - T_n(x)| &= |R_n(x)| \\ &= \left|f^{(n+1)}(c){(x-a)^{n+1}\over(n+1)!}\right| \\ &\le M^{n+1}{|x-a|^{n+1}\over(n+1)!} \\ & < M^{n+1}\left({\epsilon\over M^{n+1}}\right)\\ &= \epsilon.\end{align}$$ Hence $\displaystyle \lim_{n\to\infty}T_n(x) = f(x).$
QUESTION: I am highly doubtful that the step (!!!) is valid because there is still an $n$ in the $\epsilon$ part of the inequality. Is there something that I'm missing that would make this proof a little more safe? Or is this somewhat the correct procedure of what to do?
Step (!!!) is valid but the way you deduce it is invalid. Here is a correct way to deduce (!!!) : there is a $N_1$ such that $\left|{{|x-a|^{n+1}\over(n+1)!}\over{|x-a|^n\over n!}}\right| \leq \frac{1}{2M}$ for any $n\geq N_1$. Then , for $n > N_1$ we have $\frac{|x-a|^{n}}{n!} \leq \frac{1}{(2M)^{n-N_1}} \frac{|x-a|^{N_1}}{N_1!}$. And there is a $N_2>N_1$ such that $\frac{1}{(2M)^{n-N_2}} \leq \frac{\varepsilon}{M^{n+1}} \frac{N_1!}{|x-a|^{N_1}}$ for $n\geq N_2$. Then (!!!) follows for $n\geq N_2$.