Prove that $e>2$ geometrically.

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we see that $$ \color{#00A000}{1}+\color{#C000C0}{x}\le\left(1+\frac x2\right)^2 $$ Therefore, $$ \begin{align} 1+1 &\le\left(1+\frac12\right)^2\\ &\le\left(1+\frac14\right)^4\\ &\le\left(1+\frac18\right)^8\\ &\dots\\ &\le\lim_{n\to\infty}\left(1+\frac1{2^n}\right)^{\large2^n}\\[9pt] &=e \end{align} $$

A better (or at least alternative) definition of $e$ is this:

Let $$ L(x) = \int_1^x \frac{1}{t} dt $$ $L$ is well-defined for positive $x$ by the fundamental theorem of calculus.

With a little work, you can show that $L$ is surjective onto $R$, and since it's clearly increasing and continuous, it's also injective. So it has an inverse, $E$. $L$ is usually known as $\ln$ and $E$ is known as $\exp$.

Then $e = E(1)$ defines a new constant, called Euler's constant.

To show $e > 2$, you need only show that $L(2) < 1$. You can do this by computing an upper bound for the integral that is $L(2)$, i.e $\int_1^2 \frac{1}{t} dt$, using the partition $1, 1.5, 2$; and the left-hand ends as sample points (because $y = 1/x$ is a decreasing functions. The upper integral is then $$ \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{2}{3} = \frac{5}{6} < 1, $$ and you are done, because the integral is no larger than any of its upper integrals.