Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$.  Is there some clever way to show this other than to add the fractions together by brute-force?  For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?

$\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$
$\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}=\dfrac{5}{20}$
$\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{2}{8}=\dfrac{5}{20}$
$\dfrac{1}{9}+\dfrac{1}{11}=\dfrac{20}{99}>\dfrac{4}{20}$
