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Let $f$ be an infinitely differentiable function on an interval $I$. If $a \in I$ and there are positive constants $C$, $R$ such that for every $x$ in a neighborhood of $a$ and every $k$ it holds that

$|f^{(k)}(x)| \leq C \frac{k!}{R^k}$

then prove that the Taylor series of $f$ about $a$ converges to $f(x)$.

I think a good approach would be to estimate the error term. I'm not sure how to proceed exactly though. Thoughts?

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The way I stated it is correct. – Tony D. Mar 8 '12 at 0:27

I'm almost sure what you need for convergence is:


In such a case you would have the following:

$${R_n}\left( x \right) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^n}}}{{n!}}{f^{\left( {n + 1} \right)}}\left( t \right)dt} $$

Set $$t = x + \left( {a - x} \right)u$$

$${R_n}\left( x \right) = \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{n!}}\int\limits_0^1 {{u^n}{f^{\left( {n + 1} \right)}}\left[ {x + \left( {a - x} \right)u} \right]du} $$


$$\eqalign{ & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{n!}}{R^{n + 1}}\int\limits_0^1 {{u^n}du} \cr & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{\left( {n + 1} \right)!}}{R^{n + 1}} \cr} $$

And for $n \to \infty$ we have that $|R_n(x)| \to 0$

Here's my pick on your condition. If

$${f^{\left( {n + 1} \right)}}\left( x \right) \leqslant C\frac{{\left( {n + 1} \right)!}}{{{R^{n + 1}}}}$$

The you'd have

$$\eqalign{ & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant C \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{n!}}\frac{{\left( {n + 1} \right)!}}{{{R^{n + 1}}}}\int\limits_0^1 {{u^n}du} \cr & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant C{\left( {\frac{{\left| {x - a} \right|}}{R}} \right)^{n + 1}} \cr} $$

And the limit would be $0$ if $\left| {x - a} \right| < R$

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The second part, beginning with "Here's my pick", is the actual answer. – user147263 Jul 9 '14 at 22:46
@Thisismuchhealthier. Agreed. – Pedro Tamaroff Jul 9 '14 at 23:12

Your approach is not only correct, it's the only possible way to proceed.

Express the error term for the kth polynomial and take it's limit. How far it converges depends very heavily on the constant called $R$, but it's straightforward computation.

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