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I'm working on this analysis exercise:

"Prove that if $f$ is an infinitely differentiable function on $(a−r,a+r)$ and there is a constant $K$ such that $$\lvert f^{(n)}(x)\rvert\le K\frac{n!}{r^n}$$ for all $n\in\mathbb N$ and all $x\in(a−r,a+r)$, then the Taylor Series for $f$ at $a$ converges to $f$ on $(a−r,a+r)$."

I've tried a few things, like showing that $r$ is the radius of convergence using the ratio test, that the sequence of remainders converges to zero (using both Taylor's and Lagrange's forms for the remainder), and that the sequence of remainders is Cauchy, all to no avail. The closest I've gotten is showing that the remainders are bounded (by $\pm K$), but upon another look at the question I can't help but feel it would be easier to show if the original hypothesis were that $\lvert f^{(n)}(x)\rvert\le K$ instead of $\lvert f^{(n)}(x)\rvert\le K\frac{n!}{r^n}$, and it's happened before that the text I'm using has had errors in the statements it had wanted proven.

Thus, is this statement as originally given even true? If so, what might I be missing?

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  • $\begingroup$ It's easy to show that the power series converges uniformly to $ f$ on any $[a-s,a+s]$ for any $s$ such that $9\leq s<r.$. $\endgroup$ – DanielWainfleet Oct 30 '17 at 2:21
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No, it's right as stated. Fix an $x\in (a-r,a+r)$, and what is the estimate you get on $R_n(x)$ (the error for the $n$th Taylor polynomial) using the Lagrange formula? (Don't get confused with $x$'s here.) You should get that $|R_n(x)|\le K\left|\frac xr\right|^{n+1}$, which should be just fine to get what you want.

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  • $\begingroup$ Thanks, I must have made an error when I calculated it. I'll take another look it. $\endgroup$ – themathandlanguagetutor Oct 30 '17 at 2:19

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