¿Existe una forma fácil de ver que${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$?

Question

La suma $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ 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 $ "?

Answer 1

Dado que $y=\frac{1}{x}$ es convexo, tenemos: -

$\dfrac15+\dfrac16+\dfrac17>\dfrac36=\dfrac12$

$\dfrac18+\dfrac19+\dfrac1{10}+\dfrac1{11}+\dfrac1{12}>\dfrac5{10}=\dfrac12$

Answer 2

$\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$

$\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}=\dfrac{5}{20}$

$\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{2}{8}=\dfrac{5}{20}$

$\dfrac{1}{9}+\dfrac{1}{11}=\dfrac{20}{99}>\dfrac{4}{20}$

Answer 3

Por CS obtenemos: PS PS

Answer 4

Sin usar ningún método de fuerza bruta, una forma es observar que la suma dada es solo la diferencia de dos números armónicos :

$H_{12}-H_4= {1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} $

y $H_{4}+1<H_{12}$ .