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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 $\lim\limits_{n\rightarrow\infty}(1+\frac1n)^n=\mathrm e$, is there one (or some) alternative way(s)?

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Is this last thing your definition of $e$? – Dylan Moreland Jul 3 '12 at 21:54
up vote 44 down vote accepted

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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And from this you see that it's also at least 2.5. – Francis Adams Jul 3 '12 at 22:27
This looks like the Taylor expansion of $\exp$ at 0 evaluated at 1. Or which definition of $e$ did you use here? – krlmlr Jul 3 '12 at 22:51
@user946850, I consider $e = \sum_{n=0}^\infty 1/n!$ itself to be one of the many equivalent definitions of $e$. – Rahul Jul 3 '12 at 22:53
This shows <= (less than or equal to), doesn't it? The question asks for < (strictly less than). – ApproachingDarknessFish Dec 14 '14 at 23:36
@Approaching: There, I fixed it. – Rahul Dec 15 '14 at 3:58

You can use

$$e =\sum_{n=0}^\infty \frac{1}{n!}= 2+\sum_{n=2}^\infty \frac{1}{n!}< 2+\sum_{n=2}^\infty \frac{1}{n(n-1)}=3 \,,$$

with the last equality following immediately from the fact that $\sum_{n=2}^\infty \frac{1}{n(n-1)}$ 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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It 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 – Mikko Rantalainen May 16 '13 at 6:15
"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 at 6:52

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.

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(to show that $e>2$, integrate from 1 to 2 and show you get less than 1 by estimating with a left hand sum). – user29743 Jul 3 '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 3 '12 at 22:35
Replacing the integrand $1/t$ by its tangent at $t = (x+1)/2$ shows that $\log(x) = \int_1^x \tfrac{dt}{t} \geq 2 \tfrac{x-1}{x+1}$ for all $x \geq 1$. In particular $\log(3) \geq 1$. – WimC Jul 8 '12 at 17:46
@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. – user29743 Jul 14 '12 at 15:00

You can use integration by parts to show:

$$\int_1^e (\ln x)^2 dx = e-2$$

$$\int_1^e (\ln x )^3 dx = 6- 2e$$

Since $\ln(x)$ is strictly positive above $1$, we get

$$e-2>0$$ $$6-2e>0$$

so that $2<e<3$.

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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 $(2^1-2^0)/(1-0)= 1$. 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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Observe that, $(1 + \frac 1 n)^n = 1 + 1 + (1 - \frac 1 n)/2! + (1 - \frac 1 n)(1 - \frac 2 n)/3! + ... + (1 - \frac 1 n)(1 - \frac 2 n)...(1 - (n-1)/n)/n! < 1 + 1 + 1/2! + ... + 1/n! < 1 + 1 + 1/2 + 1/2^2 + ... + 1/2^{n-1}$, [since $n! > 2^{n-1}, \forall_n \geq 3$]. 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 $e < 3$.

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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: $\ln x_n=-(n+1)\ln (1-1/(n+1))=1+\sum_{j=1}^{\infty}(j+1)^{-1}(n+1)^{-j}.$ 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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