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Here is my candidate for the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.

$***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$

It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x > -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 < x < 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx < 0$, $1+mx < 0$ for all $m \ge n$ so certainly $(1+x)^m > 1+mx$.)

This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$ (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.

Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$. We will prove that $a_n$ is an increasing sequence and $b_n$ is an decreasing sequence. Since $a_n < b_n$, this implies, for any positive integers $n$ and $m$ with $m < n$ that $a_m < a_n < b_n < b_m$.

For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$. Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI, $((n+2)/(n+1))^{n+1} > (1+1/n)^n$, or $(1+1/(n+1))^{n+1} > (1+1/n)^n$, or $a_{n+1} > a_n$. For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$. Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI, $(n/(n+1))^{n+1} > (1-1/n)^n$ or $(1+1/n)^{n+1} < (1+1/(n-1))^n$, or $b_n > b_{n+1}$.

Since $b_n-a_n = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 < 4$, $b_n-a_n < 4/n$, so $b_n$ and $a_n$ converge to a common limit.

These proofs do not seem to be really elementary, since they use the AGMI. However, they use a special form of the AGMI, where all but one of the values are the same, and this will now be shown to be implied by BI, and thus be truly elementary.

Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive. The AGMI for these values is $(((n-1)u+v)/n)^n \ge u^{n-1}v$ with equality if and only if $u = v$. We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$ is the same as $(u+(v-u)/n)^n \ge u^n(v/u)$. Dividing by $u^n$, this is equivalent to $(1+(v/u-1)/n)^n \ge v/u$. By BI, since $(v/u-1)/n > -1/n > -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$ with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI.

marty cohen
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    What are you taking to be the definition of $e$? Also, what do you mean by elementary? Everyone I know considers the AMGM inequality elementary. – Potato May 12 '13 at 20:19
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    It does depend how you define $e$ in the first place ... I've seen it defined as the limit in your question - in which case the proof is both trivial and elementary. – Mark Bennet May 12 '13 at 20:21
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    Of course a related question is whether the limit in the title of the question exists - is that the question you were really intending to ask? – Mark Bennet May 12 '13 at 20:23
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    @Potato: So you don't know any non-mathematicians? – Harald Hanche-Olsen May 12 '13 at 20:39
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    @Potato: I consider Bernoulli's inequality more elementary than the AGMI. – marty cohen May 13 '13 at 04:05
  • In some applications you only need to know that $(1+1/n)^n $ is bounded. Is there a more elementary proof of that? – Keith McClary Sep 11 '16 at 16:14
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    @Keith McClary: If you assume the binomial theorem, you can show that $(1+1/n)^n < \sum_{k=0}^n 1/k! < 3$. However, in my opinion, the proof of the binomial theorem is itself more complicated than the proof above. – marty cohen Sep 12 '16 at 05:20
  • Relevant: https://math.stackexchange.com/questions/167843/ – Jam Oct 23 '22 at 08:08

3 Answers3

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If you just need to show that the $\lim\limits_{n\to \infty}\left(1 + \frac{1}{n}\right)^{n}$ exists, you can just show that it is an increasing sequence which is bounded above. To show that it is increasing you just need binomial theorem (an elementary result) and to prove that it is bounded, you can expand by binomial and then compare with the geometric series with ratio $\frac{1}{2}$.

Then you can define $e$ to be the limit of this sequence. And without too much effort, one may also show that

$$ e = \lim\limits_{n\to \infty}\sum\limits_{k=0}^{n}\frac{1}{k!}$$.

You will find all the details in the Principles of Mathematical Analysis by Walter Rudin.

Vishal Gupta
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Could this be ?

$$\begin{align} \frac{d}{dx} \ln x &= \lim_{h \to 0}\; \frac{\ln(x + h) − \ln x}{h} \\ &= \lim_{h \to 0} \;\frac{\ln(1 + \frac{h}{x})}{h} \\ &= \lim_{h \to 0} \;\ln\left((1 + \frac{h}{x})^{\frac{1}{h}}\right) \\ \end{align}$$

let $h=\frac xc$

$$\begin{align} \frac{d}{dx} \ln x &= \lim_{c \to \infty} \;\ln\;\left((1 + \frac{1}{c})^{\frac{c}{x}}\right) \\ &= \frac{1}{x} \;\lim_{c \to \infty}\; \ln\;\left((1 + \frac{1}{c})^{c}\right) \\ &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \end{align}$$

Since $\frac{d}{dx} \ln x = 1/x$

$$\begin{align} \frac{1}{x} &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\ 1 &= \ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\ e &= \lim_{c \to \infty}\;\left(1 + \frac{1}{c}\right)^{c} \end{align}$$

That's not very rigorous but well...

mwoua
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    if you use natural logarithm, that means you know what is $e$. That may not be the proof. – newzad May 12 '13 at 20:50
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    Nice. Not as elementary, but I'll credit you. – marty cohen May 12 '13 at 20:50
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    @newzad Napier didn't know $e$ when he defined the natural logarithm. ;-) Well, his logarithms weren't really the same as natural logarithms, but a damn good approximation of them. – egreg May 12 '13 at 20:54
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We can prove directly from BI at the expense of some algebra.

$(1+(v/u-1)/(n+1))^{n+1} > v/u$ with $u=1+1/n$ and $v=1$ is

$$\frac{1}{1+\frac{1}{n}} < \left(1+\frac{\frac{1}{1+\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right)^{n+1}$$ which gives $a_{n+1} >a_n$ in Marty's notation. With $u=1-1/n$ and $v=1$ $$\frac{1}{1-\frac{1}{n}} < \left(1+\frac{\frac{1}{1-\frac{1}{n}}-1}{n+1}\right)^{n+1} = \left( \frac{1+\frac{1}{n-1}}{1+\frac{1}{n}} \right)^{n+1}$$ Since $(1-\frac{1}{n})(1+\frac{1}{n-1}) = 1$ this gives $b_n >b_{n+1}$.

Keith McClary
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