How to show $\frac{19}{7}<e$

$$ \int_{0}^{1} x^2 (1-x)^2 e^{-x}\,dx = 14-\frac{38}{e},$$ but the LHS is the integral of a positive function on $(0,1)$.

Another chance is given by exploiting the great regularity of the continued fraction of $\coth(1)$:

$$\coth(1)=[1;3,5,7,9,11,13,\ldots] =\frac{e^2+1}{e^2-1}$$ gives the stronger inequality $e>\sqrt{\frac{133}{18}}$.

The first few convergents of the continued fraction representation of $e$ are $$ 2, 3, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \frac{106}{39} $$

Since these convergents oscillate monotonically towards $e$, the last one works to prove that $e>\frac{19}{7}$.

(If you know that $e$ is irrational, you can stop as soon as you get $\frac{19}{7}$ as a convergent because there will be other terms after that.)

This is a little easier than the OP's calculation $e\gt\sum_{n=1}^7{1\over n!}={685\over252}\gt{19\over7}$, though not by much: We can show $e^{-1}\lt{7\over19}$ via the truncation of the alternating series

$$e^{-1}\lt1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}+{1\over720}={360-120+30-6+1\over720}={265\over720}={53\over144}$$

and the cross multiplication (with some of the steps retained to make things easy to check by eye)

$$53\cdot19=1060-53=1007\lt1008=980+28=144\cdot7$$