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For all natural number $n$, $$ \frac{1}{3} + \frac{1}{5}+...+\frac{1}{2n+1}$$ is never an integer.

I know that $\sum_{k=1}^{n}\frac{1}{k}$ is never an integer, but I'm not sure how to restrict that to only odd denominators.

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    $\begingroup$ $$\sum_{k=1}^{2n+1}\dfrac1k=\dfrac12\sum_{k=1}^n\dfrac1k+\sum_{k=0}^n\dfrac1{2k+1}$$ $$\implies \sum_{k=0}^n\dfrac1{2k+1}=?$$ $\endgroup$ – lab bhattacharjee Mar 5 '18 at 5:53
  • $\begingroup$ I edited my question, the series only includes odd denominators. :) $\endgroup$ – wtnmath Mar 5 '18 at 5:56
  • $\begingroup$ math.stackexchange.com/questions/2019926/… $\endgroup$ – Andrew Li Mar 5 '18 at 6:00
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    $\begingroup$ @lab bhattacharjee Thanks! But how is $\frac{1}{2}(2H_{2n}−H_n)$ for sure not an integer given that $2H_{2n}$ and $H_n$ aren't. $\endgroup$ – wtnmath Mar 5 '18 at 7:10
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Given $n$, let $m$ be the positive integer such that $3^m \leqslant 2n+1 < 3^{m+1}$. Then in

$$\sum_{k = 1}^{n} \frac{1}{2k+1}$$

there is only one term whose denominator is divisible by $3^m$, namely $\frac{1}{3^m}$ (since $2\cdot 3^m$ is even, and $r\cdot 3^m > 2n+1$ for $r \geqslant 3$). Thus if

$$\Biggl(\sum_{k = 1}^n \frac{1}{2k+1}\Biggr) - \frac{1}{3^m} = \frac{a}{b}$$

with $a$ and $b$ coprime positive integers (or $a = 0$ in the case $n = 1$), then $3^m$ does not divide $b$. Write $b = 3^s\cdot c$ with $c$ not divisible by $3$. Then

$$\sum_{k = 1}^{n} \frac{1}{2k+1} = \frac{a}{3^s\cdot c} + \frac{1}{3^m} = \frac{3^{m-s}\cdot a + c}{3^m\cdot c},$$

and since $m > s$ the numerator is not divisble by $3$, so the fraction is not an integer.

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