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?
Best Answer
I'm almost sure what you need for convergence is:
$$|f^{(n)}(x)|<R^n$$
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 $$t = x + \left( {a - x} \right)u$$
$${R_n}\left( x \right) = \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{n!}}\int\limits_0^1 {{u^n}{f^{\left( {n + 1} \right)}}\left[ {x + \left( {a - x} \right)u} \right]du} $$
Then
$$\eqalign{ & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{n!}}{R^{n + 1}}\int\limits_0^1 {{u^n}du} \cr & 0 \leqslant \left| {{R_n}\left( x \right)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{\left( {n + 1} \right)!}}{R^{n + 1}} \cr} $$
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 $\left| {x - a} \right| < R$