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
 A: There is one essential mistake in your question (your lecture notes, however, did it correctly): If $R$ is the radius of convergence of a power series
\begin{align*}
f(z):=\sum_{k=0}^\infty a_k(z-z_0)^k,
\end{align*}then it converges for all $|z-z_0|\color{red} < R$. Furthermore, for each $0<r\color{red}<R$ we have that
\begin{align*}
|a_k|\leq\frac{M_f(r)}{r^k},\qquad k\in\mathbb N_0,
\end{align*}
where
\begin{align*}
M_f(r):=\sup_{|z-z_0|=r}|f(z)|.
\end{align*}
In particular, since $a_k=\frac{f^{(k)}(z_0)}{k!}$ it follows that
\begin{align*}
|f^{(k)}(z_0)|\leq M_f(r)\cdot\frac{k!}{r^k},\qquad k\in\mathbb N_0.
\end{align*}
Now, in your notes you can assume that the radius of convergence is larger that $R$, hence the power series converges for all $|x|\color{red}\leq R$ and your statement is correct using $C=M_f(R)$.
A: 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 ($R=1$)
$$
\lvert\,f^{(n)}(0)\rvert = n\cdot \frac{n!}{R^n}.
$$
