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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 $1$"?

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  • $\begingroup$ On my calculator the sum is $1.019877344$. $\endgroup$ Commented Jan 19, 2020 at 0:18
  • $\begingroup$ @OscarLanzi Yes, I know what the sum is. I'm looking for a way to recognize that it is larger than 1 without actually adding all of the numbers together. $\endgroup$
    – mweiss
    Commented Jan 19, 2020 at 0:19

4 Answers 4

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For positive, unequal $a$ and $b$:

$\dfrac1a+\dfrac1b=\dfrac{a+b}{ab}>\dfrac4{a+b}$

because $(a+b)^2>4ab$ (the difference between these is $(a-b)^2$). So,

$\dfrac15+\dfrac17>\dfrac4{12}=\dfrac13$

$\dfrac19+\dfrac1{11}>\dfrac4{20}=\dfrac1{5}$

$\dfrac18+\dfrac1{12}>\dfrac4{20}=\dfrac1{5}$

When these inequalities are put into the given sum the claimed bound follows.

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    $\begingroup$ This is pretty much exactly what I was looking for. Thanks! $\endgroup$
    – mweiss
    Commented Jan 19, 2020 at 0:16
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    $\begingroup$ You could mention that this is the inequality between harmonic and arithmetic mean: $\frac{2}{1/a+1/b} < \frac{a+b}{2}$. $\endgroup$
    – Martin R
    Commented Jan 20, 2020 at 7:55
  • $\begingroup$ @mweiss if this solution is what you are looking for, then I think it should be chosen as an answer. $\endgroup$ Commented Jan 29, 2020 at 3:18
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Since $y=\frac{1}{x}$ is convex we have:-

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

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

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  • $\begingroup$ This is also great. Thank you! $\endgroup$
    – mweiss
    Commented Jan 19, 2020 at 1:17
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$\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$

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

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

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

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By C-S we obtain: $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}=$$ $$=\frac{17}{60}+\frac{17}{66}+\frac{17}{70}+\frac{17}{72}\geq\frac{17\cdot4^2}{60+66+70+72}=\frac{68}{67}>1.$$

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    $\begingroup$ My analysis uses C-S with pairs of terms. +1 for getting a bound actually greater than $1$. $\endgroup$ Commented Jan 19, 2020 at 12:10

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