TITLE: Show that $\left(1+\dfrac{x}{n}\right)^n \to e^x$ as $n \to \infty$ in a normed ring $R$
QUESTION [10 upvotes]: I had an interesting discussion yesterday with one of my friends (I think he is a member here, am I right?). He claimed that
$$\left(1+\dfrac{x}{n}\right)^n \to e^x$$
basically in any normed ring $R$ (with a copy of $\mathbb{Q}$, am I right?) as
$$n \to \infty.$$
Unfortunately, he only proved it for partially ordered rings, and his proof looked suspiciously like a certain Wikipedia article. I tried to replace everything with norms, but I only ended up with a big mess. I think I've missed the key idea here, am I right?
I also wonder if the conditions can be relaxed further, but I don't think it is possible. Am I right?

REPLY [11 votes]: Let's assume (as Landscape does) that we are in a Banach algebra. By the binomial theorem, one may write
\begin{equation} \left(1+\frac{x}n\right)^n=\sum_{k=0}^\infty a_{k,n} \, ,\tag{$*$}\end{equation}
where the $a_{k,n}$ are given by 
$$a_{k,n}=\left\{
\begin{matrix}
\left(n\atop k\right) \frac{x^k}{n^k}&{\rm if}\; k\leq n\\
0&{\rm if}\; k>n
\end{matrix}\right.$$
For each fixed $k$, we have $\lim_{n\to\infty} a_{k,n}=\frac{x^k}{k!}$, because $\left(n\atop k\right)=\frac{n(n-1)\cdots (n-k+1) }{k!}\sim\frac{n^k}{k!}$. Moreover, we also have $\left(n\atop k\right)\leq \frac{n^k}{k!}$, so that $\Vert a_{k,n}\Vert\leq \frac{\Vert x\Vert^k}{k!}\cdot$ Since the series $\sum\frac{\Vert x\Vert^k}{k!}$ is convergent, we may apply the "dominated convergence theorem for series" to let $n\to\infty$ in $(*)$. This gives 
$$\lim_{n\to\infty} \left(1+\frac{x}n\right)^n=\sum_{k=0}^\infty \frac{x^k}{k!}=e^x\, .$$