How to show $\frac{19}{7}How can I show $\dfrac{19}{7}<e$ without using a calculator and without knowing any digits of $e$?
Using a calculator, it is easy to see that $\frac{19}{7}=2.7142857...$ and $e=2.71828...$ 
However, how could this be shown in a testing environment where one does not have access to a calculator?
My only thought is to use the Taylor series for $e^x$ with $x=1$ to calculate $\displaystyle e\approx\sum \limits_{n=0}^{7}\frac{1}{n!}=\frac{685}{252}=2.7182...$ 
However, this method seems very time consuming and tedious, finding common denominators and performing long division. Does there exist a quicker, more elegant way?

Since you know what the result should be, try remove terms one by one by computing residues $r_i$. That will reduce the size of integers, as long as yoou factorize while you can, which is easy because of the factorials (highly composite numbers). 
First, notice that the first $3$ terms for $e$ give $\frac{5}{2}$.
 Now, find residues:
$$r_1 = \frac{19}{7} - \frac{5}{2} = \frac{3}{2.7}\,, $$
$$r_2 = \frac{3}{2.7} -\frac{1}{6} = \frac{1}{2}\left(\frac{3}{7}-\frac{1}{3}\right) =\frac{1}{3.7}\,,$$
$$r_3 =\frac{1}{3.7} -\frac{1}{3.8} = \frac{1}{3}\frac{1}{7.8}\,,$$
$$r_4 =\frac{1}{3.7.8} -\frac{1}{120} = \frac{1}{3.8}\left(\frac{1}{7}-\frac{1}{5}\right)\,.$$
Since $5<7$, the last residue $r_4$ is negative, you can stop here, never having to do long multiplications, the hardest being $19\times 2$.
