Proving $\mathrm e <3$

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)?

Solution 1:

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: $$\begin{align} &\quad 1 + 1 &= 2\\ &< 1 + 1 + \frac12 + \frac1{2\cdot3} + \frac1{2\cdot3\cdot4} + \cdots &= e \\ &< 1 + 1 + \frac12 + \frac1{2\cdot2} + \frac1{2\cdot2\cdot2} + \cdots &= 3 \end{align}$$

Solution 2:

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 $2<e<3$.

Solution 3:

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

$$\frac{1}{3} < \frac{1}{e} < \frac{1}{2} \,.$$