Convergence of the $e^x$ Taylor series

We have an amazing thing called Lagrange remainders. They basically tell us the difference between our function and it's Taylor polynomial. In general, we have

$$R_n(x)=|f(x)-P_n(x)|$$

where $P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$. Since it follows that

$$R_n(a)=0\\R_n'(a)=0\\R_n''(a)=0\\\vdots\\R_n^{(n)}(a)=0\\R_n^{(n+1)}(a)=|f^{(n+1)}(a)|$$

Thus,

$$R_n^{(n+1)}(x)\le|f^{(n+1)}(c)|$$

for some $c$ in our radius of convergence. It thus follows by integrating a few times that

$$R_n(x)\le\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$

One can then see that as $n\to\infty$, we have

$$|f(x)-P(x)|\le\lim_{n\to\infty}\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$

and if $\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\to0$ for any $x,c$ within the a given domain, then the power series will equal the original function over that domain.

See if you can show that for any $x,c\in\mathbb R$,

$$\lim_{n\to\infty}\left|\frac{e^c}{(n+1)!}x^{n+1}\right|=0$$

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On a side note, Lagrange remainder also shows us how well we approximate something when using a power series. For example, if I wanted to calculate $e$ out 5 places accurately,

$$R_n(x)=\left|e^x-\sum_{k=0}^n\frac{x^n}{n!}\right|\le0.000001$$

It's easy enough to solve, since

$$R_n(x)\le\left|e\frac{x^{n+1}}{(n+1)!}\right|\le\left|3\frac{x^{n+1}}{(n+1)!}\right|$$

Our particular case is $x=1$, and thus it suffices to solve

$$\frac3{(n+1)!}<0.000001$$

Which is easily done with a few checks to give $n\le8$. Thus,

$$e=\pm0.000001+\sum_{k=0}^8\frac1{k!}$$

You can do this in steps:

1. Define the function $\exp x = \sum_{j=0}^\infty x^j/j!$. Show that it converges everywhere.

2. Using Cauchy products and the Binomial Theorem, show that $\exp (w + z) = \exp w \exp z$.

3. Show that $\exp$ is continuous and increasing. Differentiation gives $\exp' = \exp$, and $\exp$ is positive everywhere.

4. Since you know $\exp 1 = \mathrm e$, an inductive application of (2) gives $$ (\exp q)^m = \exp(mq) = \exp n = \mathrm e^n $$ for any rational $q=n/m$, so $\exp q = \mathrm e^q$.

5. Via the definition of power $\mathrm e^x = \sup_{q \leq x} \mathrm e^q$ and (3), show that $\exp x = \mathrm e^x$ for all $x$.