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Possible Duplicate:
Is there an elementary proof that $∑_{k=1}^n 1/k$ is never an integer?

Hello,

Prove that $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$ is not an integer.

I tried to prove by induction on $n$, but I was stuck :(

Assume $1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} = \frac{a}{b}$ for some integers $a, b$ and $a \neq b \text{and} b \neq 0$
Then $ 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n + 1} = \frac{a}{b} + \frac{1}{n + 1}$

Then how can I prove that this expression is not integer? A hint would be greatly appreciated.

Thanks,
Chan

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    $\begingroup$ Look at the power of 2 divisible by numerator and denominator. $\endgroup$
    – user325
    Feb 28, 2011 at 0:24
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    $\begingroup$ Your induction hypothesis is not strong enough, because simply assuming that $k$ is not an integer does not guarantee that $k+\frac{1}{n+1}$ is not an integer. So if you want to proceed by induction, you need to prove more than simply that $H_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$ is not an integer, you need to prove something about its expression as a rational written in lowest terms. $\endgroup$
    – Arturo Magidin
    Feb 28, 2011 at 0:27
  • $\begingroup$ @Arturo Magidin: Thanks, that was exactly the problem that I encountered, since I actually found counter examples. $\endgroup$
    – roxrook
    Feb 28, 2011 at 0:29
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    $\begingroup$ Related: math.stackexchange.com/questions/2746/… and math.stackexchange.com/questions/5219/… $\endgroup$
    – Jonas Meyer
    Feb 28, 2011 at 0:31
  • $\begingroup$ Just out of curiosity. Would saying that for $n=2$ the series is not an integer suffice? Proof by example? $\endgroup$
    – Jacob
    Feb 28, 2011 at 0:37

2 Answers 2

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HINT: There is always a prime between $\frac{n}{2}$ and $n$, $\forall n \geq 4$

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    $\begingroup$ That's overkill. $\endgroup$
    – lhf
    Feb 28, 2011 at 0:39
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Hint: look at the largest power of 2 less than $n$. Can it get canceled out from the denominator?

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