# Non-Rigorous Arguments 1: Two Formulas For e

I’m a big fan of non-rigorous arguments, especially in calculus and analysis. I think there should be a book cataloging all the beautiful, morally-true-but-not-actually-true proofs that mathematicians have advanced, but until that time I’ll try to at least catalog a few of them on my blog.

This first one is Euler’s original argument for the equality of two expressions (both of which happen to define $e$): $\displaystyle{\sum_{n=0}^\infty \frac{1}{n!} = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n}$

I’ll also sketch how this can be made rigorous in non-standard analysis.

The argument is as follows: The limit $\lim_{n\to\infty} (1 + 1/n)^n$ is equal to $(1 + 1/N)^N$, where $N$ is infinitely large. By the binomial theorem, this is: $\displaystyle{1 + N\frac{1}{N} + {N\choose 2}\frac{1}{N^2} + \cdots + \frac{1}{N^N} = \sum_{i=0}^N {N\choose i}\frac{1}{N^i}}$

Since $N\choose i$ is $N(N-1)\cdots (N- (i - 1))/ i!$, this is the sum as $i$ ranges from ${0}$ to $N$ of: $\displaystyle{\frac{N(N-1)\cdots (N-(i-1))}{N^i}\cdot\frac{1}{i!}}$

Now, if $i$ is infinitely large, this term is so small that it may be neglected. On the other hand, if $i$ is finite, then $N - j = N$ for $j< i$. Therefore $\displaystyle{\frac{N(N-1)\cdots (N-(i-1))}{N^i}=1}$

and the whole sum is equal to $\displaystyle {\sum_{i=0}^N \frac{1}{i!} = \sum_{i=0}^\infty \frac{1}{i!}}$,

as desired.

Now, I’ll sketch how to make this rigorous in non-standard analysis. This is from Higher Trigonometry, Hyperreal Numbers, and Euler’s Analysis of Infinities by Mark McKinzie and Curtis Tuckey, which is the best introductory article on non-standard analysis that I’ve read.

In non-standard analysis, one extends the real numbers $\mathbb{R}$ to a larger field $\mathbb{R}^*$ which contains all the reals, but also a positive $\epsilon$ which is less than every positive real (and hence also a number $1/\epsilon$ which is greater than every real). For every function $f\colon \mathbb{R}^n \to \mathbb{R}$, there is a function $f^*\colon (\mathbb{R}^*)^n \to \mathbb{R}^*$, and the $f^*$‘s satisfy all the same identities and inequalities formed out of composition that the $f$‘s do. (For example, $\sin^*(2x) = 2\sin^*(x)\cos^*(x)$ for all hyperreal $x$.) For that reason, I’ll often omit the ${}^*$. The range of $(x \mapsto \lfloor x \rfloor)^*$ is called $\mathbb{Z}^*$, the hyperintegers. Since $x - 1\leq \lfloor x \rfloor \leq x$, the same is true in the hyperreals and there are therefore infinite hyperintegers.

We call a nonzero hyperreal $x\in \mathbb{R}^*$ infinitesimal if $|x|$ is less than every positive real. We say that $x$ and $y$ are close (written $x\simeq y$) if $x - y$ is infinitesimal. We say that $x$ is infinite if $1/x$ is infinitesimal (equivalently, if $|x|$ is greater than every real). We say that $x$ is finite if it’s not infinite (equivalently, if $|x|$ is less than some real).

Let $\{s_n\}_{n=0}^\infty$ be a sequence of hyperreals. We say that it is determinate if $\sum_{n=0}^N s_n \simeq \sum_{n=0}^M s_n$ whenever $N$ and $M$ are infinite

The Summation Theorem can then be proven: If $\{s_n\}_{n=0}^\infty$ and $\{t_n\}_{n=0}^\infty$ are two determinate sequences such that $s_n \simeq t_n$ for all finite $n$, then $\sum_{n=0}^N s_n \simeq \sum_{n=0}^N t_n$ for all infinite $N$.

By appropriately replacing “equals” with “is close to”, Euler’s argument above may now be adapted to prove that for all infinite $N$ and $M$, $\displaystyle{\left(1 + \frac{1}{N}\right)^N \simeq \sum_{n=0}^M \frac{1}{n!}}$

(the sequence $s_n = 1/n!$ may be proved determinate by comparison with the geometric sequence, which is easily shown determinate). By a transfer principle, this may in turn be used to prove that $\lim_{n\to \infty} (1 + 1/n)^n = \sum_{n=0}^\infty 1/n!$ (in the regular reals).

## 2 thoughts on “Non-Rigorous Arguments 1: Two Formulas For e”

1. vahid says:

may i ask is there any formula for calculating e as fast as possible ! which is the fastest formula?

2. mathphile says:

Try ( 1+ 1/100,000,000 ) ^ 100, 000, 000 on your calculator.

This is accurate to 7 decimal places. Nothing special. You could use many other series representations for e. Accuracy of 7 digits is more than sufficient for all calculations on planet earth.

The magnitude e is always used as a rational number even though it is not a number at all.