Uniform Convergence Involving $e^x$ my question is:
Consider the Taylor polynomial of $e^x$ of order $n$ at $x=0$, or $P_n(x)=e^0 + e'^{(0)}x +...+\frac{e^{(n)(0)}x^n}{n!}$.
Prove that the sequence {$P_n(x)$} converges uniformly to $e^x$ on any bounded interval.
I tackled the first part of the question by proving, through Taylor's theorem, that $e^x = P_n(x) +$ some error term, and I've simplified the epsilon expression to be $\frac{e^cx^{n+1}}{(n+1)!}<\epsilon$. I'm trying to create a $k>n$ out of this, but I'm not sure of what I should do. I thought of showing $(n+1)!\ge n$ to make $\frac{e^cx^{n+1}}{(n+1)!} \le \frac{e^cx^{n+1}}{n} < \epsilon$ , so I can swap the $n$ with the $\epsilon$, but apparently that's not right. Thanks in advance for your help.
 A: We have to show that for any $\epsilon\gt0$ and $M\gt0$ there is an $N$ so that if $n\ge N$ and $|x|\le M$, then
$$
\left|e^x-\sum_{k=0}^n\frac{x^k}{k!}\right|\le\epsilon
$$
We can achieve this by showing
$$
\sum_{k=n+1}^\infty\frac{|x|^k}{k!}\le\epsilon
$$
For $k\ge2M$, $\frac{M^k}{k!}=\frac{M^{2M}}{(2M)!}\overbrace{\frac{M}{2M+1}\frac{M}{2M+2}\cdots\frac{M}{k}}^{k-2M\text{ terms}}\le\frac{M^{2M}}{(2M)!}\frac1{2^{k-2M}}=\frac{(2M)^{2M}}{(2M)!}2^{-k}$.
By choosing $N\ge\log_2\left(\frac1\epsilon\frac{(2M)^{2M}}{(2M)!}\right)$, we get for $n\ge N$ and $|x|\le M$,
$$
\begin{align}
\sum_{k=n+1}^\infty\frac{|x|^k}{k!}
&\le\sum_{k=n+1}^\infty\frac{M^k}{k!}\\
&\le\sum_{k=n+1}^\infty\frac{(2M)^{2M}}{(2M)!}2^{-k}\\
&=\frac{(2M)^{2M}}{(2M)!}2^{-n}\\[9pt]
&\le\epsilon
\end{align}
$$

We can also use the error term $e^x-P_n(x)=\dfrac{e^\xi x^{n+1}}{(n+1)!}$ for some $\xi$ between $0$ and $x$. Given an $\epsilon\gt0$ and $M\gt0$, we want to find an $N$ so that if $n\ge N$ and $|x|\le M$, then
$$
\frac{e^\xi x^{n+1}}{(n+1)!}\le\epsilon
$$
For $k\ge2M$, $\frac{M^k}{k!}=\frac{M^{2M}}{(2M)!}\overbrace{\frac{M}{2M+1}\frac{M}{2M+2}\cdots\frac{M}{k}}^{k-2M\text{ terms}}\le\frac{M^{2M}}{(2M)!}\frac1{2^{k-2M}}=\frac{(2M)^{2M}}{(2M)!}2^{-k}$.
By choosing $N\ge\log_2\left(\frac{e^M}\epsilon\frac{(2M)^{2M}}{(2M)!}\right)$, we get for $n\ge N$ and $|x|\le M$ and $\xi$ between $0$ and $x$,
$$
\begin{align}
\frac{e^\xi x^{n+1}}{(n+1)!}
&\le\frac{e^MM^{n+1}}{(n+1)!}\\
&\le e^M\frac{(2M)^{2M}}{(2M)!}2^{-n-1}\\[9pt]
&\le\epsilon
\end{align}
$$
A: The key to this proof is that for a fixed $x$, $x^n\leq n!$ when $n$ is sufficiently large. Let's consider the ratio 
$$\frac{x^n}{n!}=\frac{x\cdot x\cdot\ldots\cdot x}{1\cdot 2\cdot 3\cdot\ldots\cdot (n-1)\cdot n}$$
If $x$ is large, then for small $n$, the numerator is clearly larger. However, once $n\geq x$, each successive term in the numerator are the same (we just multiply by $x$), but successive terms in the denominator are still growing, so the denominator starts winning. For simplicity I'll assume $x$ is an integer, but it's not necessary (just take the floor of $x$). We can make this precise by assuming $n\geq x$ and then rewriting the expression as the fir
$$ \frac{x^n}{n!}=\frac{x^x}{x!}\frac{x^{n-x}}{n\cdot (n-1)\cdot\ldots\cdot(x+1)}$$
The first expression is a constant, and the second expression can be made as small as we like, for example by (badly) estimating it by $\left(x/(x+1)\right)^{n-x}$.
