Well I am just asking myself if there's a more elegant way of proving $$2<\exp(1)=\mathrm e<3$$ than doing it by induction and using the fact of $\lim\limits_{n\rightarrow\infty}\left(1+\frac1n\right)^n=\mathrm e$, is there one (or some) alternative way(s)?

6Is this last thing your definition of $e$? – Dylan Moreland Jul 03 '12 at 21:54
7 Answers
What answer you find most elegant may depend on what definition of $e$ you're starting with, as Dylan suggests, but I find this argument quite short and sweet: $$\begin{align} &\quad 1 + 1 &= 2\\ &< 1 + 1 + \frac12 + \frac1{2\cdot3} + \frac1{2\cdot3\cdot4} + \cdots &= e \\ &< 1 + 1 + \frac12 + \frac1{2\cdot2} + \frac1{2\cdot2\cdot2} + \cdots &= 3 \end{align}$$

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This looks like the Taylor expansion of $\exp$ at 0 evaluated at 1. Or which definition of $e$ did you use here? – krlmlr Jul 03 '12 at 22:51

6@user946850, I consider $e = \sum_{n=0}^\infty 1/n!$ itself to be one of the many equivalent definitions of $e$. – Jul 03 '12 at 22:53

I don't understand the "$1 + 1 + \frac12 + \frac1{2\cdot2} + \frac1{2\cdot2\cdot2} + \cdots = 3$" part. Is there a formula to get it? – yaobin May 18 '19 at 01:36

3@yaobin Yes, there is. This sum is $1 + \sum_{k = 0}^{\infty} 2^{k}$ which is a geometric series with $r = \frac{1}{2}$. Therefore, $$\sum_{k = 0}^{\infty} 2^{k} = \frac{1}{1  \frac{1}{2}} = 2.$$ – Ramanujan Jun 03 '19 at 18:00
You can use integration by parts to show:
$$\int_1^e (\ln x)^2 dx = e2$$
$$\int_1^e (\ln x )^3 dx = 6 2e$$
Since $\ln(x)$ is strictly positive above $1$, we get
$$e2>0$$ $$62e>0$$
so that $2<e<3$.
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4By the way, this argument left alone might look a bit adhoc, but it generalizes: integrating $\int_1^e (\ln x)^n dx$ by parts shows that $\frac{n!}{!n}$ alternates around $e$. This sequence actually converges extremely fast, so you can use it pretty easily to compute arbitrary precision rational approximations to $e$. – Marcel Besixdouze Oct 20 '18 at 21:42

1Actually, $\int_1^e (\ln x)^n dx = a_n e + b_n$, where $(1)^n a_n$ is https://oeis.org/A000166 and $(1)^{n+1} b_n=n!$. – lhf Jan 10 '20 at 10:07
You can use
$$e =\sum_{n=0}^\infty \frac{1}{n!}= 2+\sum_{n=2}^\infty \frac{1}{n!}< 2+\sum_{n=2}^\infty \frac{1}{n(n1)}=3 \,,$$
with the last equality following immediately from the fact that $\sum_{n=2}^\infty \frac{1}{n(n1)}$ is telescopic.
Of course it depends on the way you define $e$, anyhow the equality
$$\sum_{n=0}^\infty \frac{1}{n!}=\lim\limits_{n\rightarrow\infty}(1+\frac1n)^n$$ can be established easily using the binomial theorem.
Second solution
You can use the fact that $a_n=(1+\frac{1}{n})^{n+1}$ is decreasing. The inequality $a_{n+1} < a_n$ is an immediate consequence of Bernoulli Inequality.
Note that this implies (induction hidden here) that $a_n \leq a_6 <3$ for all $n \geq 3$, and that
$$e =\lim a_n \leq a_6 <3 \,.$$
Here is one more:
$$e^{1}=1\frac{1}{1!}+\frac{1}{2!}\frac{1}{3!}+.. \,.$$
Since the series is alternating and $\frac{1}{n!}$ is decreasing, it is obvious (very easy to show) that the series oscilates around the limit and
$$s_{2n+1}=1\frac{1}{1!}+\frac{1}{2!}\frac{1}{3!}+....+\frac{1}{(2n)!}\frac{1}{(2n+1)!} \leq \frac{1}{e} \leq 1\frac{1}{1!}+\frac{1}{2!}\frac{1}{3!}+....+\frac{1}{(2n)!}=s_{2n}$$
[ Actually in the proof of the Alternating series test, one proves the stronger statement that for such a series we have $s_{2n}$ decreasing, $s_{2n+1}$ increasing and $s_{2n+1} \leq s_{2n}$. ]
The inequality
$$1\frac{1}{1!}+\frac{1}{2!}\frac{1}{3!} < \frac{1}{e} < 1\frac{1}{1!}+\frac{1}{2!}$$ is
$$\frac{1}{3} < \frac{1}{e} < \frac{1}{2} \,.$$
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1It was not clear to me how to easily show using binomial theorem that both definitions of $e$ are equal. Fortunately, Wikipedia had the answer at http://en.wikipedia.org/wiki/Binomial_theorem#Series_for_e – Mikko Rantalainen May 16 '13 at 06:15

2"Note that this implies (induction hidden here) that $a_n\le a_6<3$ for all $n\ge 3$" — I guess you mean $n\ge 6$ here. – celtschk Jul 23 '16 at 06:52
It's equivalent to show that the natural logarithm of 3 is bigger than 1, but this is $$ \int_1^3 \frac{dx}{x}. $$ A right hand sum is guaranteed to underestimate this integral, so you just need to take a right hand sum with enough rectangles to get a value larger than 1.

(to show that $e>2$, integrate from 1 to 2 and show you get less than 1 by estimating with a left hand sum). – Jul 03 '12 at 22:09

You can also use the fact that $\int_1^3 \frac{1}{x^a} >1$ when $a$ is close enough to $1$. I think it holds for $a=1.1$, but that is not easy to show.... – N. S. Jul 03 '12 at 22:35

3Replacing the integrand $1/t$ by its tangent at $t = (x+1)/2$ shows that $\log(x) = \int_1^x \tfrac{dt}{t} \geq 2 \tfrac{x1}{x+1}$ for all $x \geq 1$. In particular $\log(3) \geq 1$. – WimC Jul 08 '12 at 17:46

1@WimC, wow, graphing software shows that is an outstanding estimate between .5 and 1.5, had not seen that one before. That's really cool. – Jul 14 '12 at 15:00
First let's consider a simple heuristic argument to show that $2<e<4$. It is easy to prove using the definition of the derivative that if $f(x)=2^x$ then $f'(x) = (\text{constant}\cdot 2^x$). The curve gets steeper as $x$ increases, and the average slope between $x=0$ an $x=1$ is $(2^12^0)/(10)= 1$. Therefore, the slope at $x=0$ is less than $1$; hence the "constant" is less than $1$. Now do the same with $g(x)=4^x$ on the interval from $x=1/2$ and $x=0$, and conclude that the slope at $x=0$ is more than $1$; hence the "constant" you get there is more than $1$.
So $2$ is too small, and $4$ is too big, to serve as the base of the natural exponential function.
It's messier to do the same with $3$, but the interval from $x=1/6$ to $x=0$ will do it, and you conclude $3$ is too big to be the base of the natural exponential function.
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The sequence $x_n=(1+1/n)^{n+1}$ is strictly decreasing and converges to $e.$ So $e<x_5=2.985984 <3.$
Remark: $\ln x_n=(n+1)\ln (11/(n+1))=1+\sum_{j=1}^{\infty}(j+1)^{1}(n+1)^{j}.$ Comparing the terms of this series to the corresponding terms in the series for $\ln x_{n+1} , $ we see that $\ln x_n>\ln x_{n+1} , $ so $ x_n>x_{n+1}.$
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Observe that, $(1 + \frac 1 n)^n = 1 + 1 + (1  \frac 1 n)/2! + (1  \frac 1 n)(1  \frac 2 n)/3! + ... + (1  \frac 1 n)(1  \frac 2 n)...(1  (n1)/n)/n! < 1 + 1 + 1/2! + ... + 1/n! < 1 + 1 + 1/2 + 1/2^2 + ... + 1/2^{n1}$, [since $n! > 2^{n1}, \forall_n \geq 3$]. So, $(1 + \frac 1 n)^n < 3  (\frac 1 2)^n$.Now as $n$ is tending to infinity $(\frac 1 2)^n$ diminishes indefinitely and ultimately we obtain $e < 3$.