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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?

Anonymous - a group
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    Generally, you seem to be right. – Kortlek Jun 05 '13 at 20:41
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    What does "normed ring" mean in your context? Can I assume that $R$ is a Banach algebra? – 23rd Jun 11 '13 at 19:22
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    Justen Bieber, Denis Rodman, KimJung-Un, Obama, Romney, etc...what a job to maintain all these accounts active simultaneously. And what a dedication to your joke. – Julien Jun 14 '13 at 22:46
  • @DenisRodman: OK. Anyway, Etienne has already provided an answer under this setting. – 23rd Jun 15 '13 at 06:16

1 Answers1

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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 $$\lim_{n\to\infty} \left(1+\frac{x}n\right)^n=\sum_{k=0}^\infty \frac{x^k}{k!}=e^x\, .$$

Etienne
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