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 $x,c\in\mathbb R$,

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

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 $x$.

First two points about your question:

• You say that you have figured out that $e=\sum 1/i!$. This suggests that you are using a definition of $e$ which is different from that sum. So how do you define $e$?
• Next is how do you define $e^{x} $? If $x$ is rational then a definition of $e$ is sufficient. For irrational $x$ one must define the concept of irrational exponent and this is a non-trivial task. How do you do that?

There are many ways to answer both the questions above and the simplest way is perhaps to use the definition $$\exp(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}\tag{1}$$ and then define $e=\exp(1)$. It can be proved that the definition above makes sense (that is the limit in $(1)$ exists). Further we can prove that $\exp(x+y) =\exp(x) \exp(y) $ and then by algebra we get $\exp(x) =e^{x} $ for all rational $x$.

Next step is to define irrational exponent $x$ at least for the base $e$ by the equation $e^{x} =\exp(x) $ and deal with symbol $a^{x}$ for general $a$ later.

The result $$\exp(x) =\sum_{i=0}^{\infty}\frac{x^{i}}{i!}\tag{2}$$ can be proved using $(1)$ with some effort. I have given details of this approach in my blog post.

If your objective is to establish $(2)$ using Taylor's theorem then you need to show that derivative of $\exp(x) $ is $\exp(x) $ itself. This is already presented in this answer.