Upper Bound for $|f^{n}(0)|$ given that $f$ is Analytic Let $f(x)$ be an analytic function in some neighborhood of $x=0$.
$f$ being analytic implies that its has a convergent Taylor series expansion about $x=0$. That is, there exists $R>0$ (radius of convergence) such that the Taylor series $$\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^{n}$$ for all $|x|\leq R$.
The following is the part I do not understand. Any help will be appreciated.

This implies the existence of a constant $C$ such that $$|f^{n}(0)|\leq C\frac{n!}{R^n}.$$

Is it correct to say that for a convergent Taylor series, each term has an upper bound, say $C$, such that $\left|\frac{f^{n}(0)}{n!}x^{n}\right|\leq C$? 
Added Later. The question is based on the following online notes (page 2, equations 10 and 11): Notes

This is not true. 
Let $f(x)=\sum_{n=0}^\infty  nx^n$. Then $f$ is analytic in $(-1,1)$, and the power series has radius of convergence, around zero, $R=1$. But
$$
f^{(n)}(0)=n\cdot n!,
$$ 
and hence ($R=1$)
$$
\lvert\,f^{(n)}(0)\rvert = n\cdot \frac{n!}{R^n}.
$$
