Prove that :

\[1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} (\forall n \in \mathbb{Z}) \]

Prove :

Assume \(1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} \)

So that : \(1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} = \frac{a}{b} ; (a,b) = 1 \)

We must demonstrate : \(1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} \notin \mathbb{Z} \)

Pretend that : \(1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} \in \mathbb{Z} \)

\( \Rightarrow \frac{a}{b} + \frac{1}{n + 1} \in \mathbb{Z} \) \( \Rightarrow \frac{a(n + 1) + b}{b(n + 1)} \in \mathbb{Z} \) \( \Rightarrow a(n + 1) + b \vdots b(n + 1) \)

By (a,b) = 1 then \( n + 1 \vdots b \) and \( b \vdots n + 1 \) so \(n+1 = b\)

But :

\[1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} = \frac{a}{n+1} = \frac{c}{n!} \] Thus, \[ n! \vdots n+1 ;(absurdity) \]

Therefore : \[1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} (\forall n \in \mathbb{Z}) \]

Thanks in advance ~!!

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## Comments

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TopNewestGood attempt. However, there is an error in the last line. So what if we have \( \frac{a}{ n+1} = \frac{c}{n!} \)? Why does that mean that \( n! \mid n+1 \)? For example, \( \frac{2}{4} = \frac{ 3}{6} \), and 4 and 6 do not divide each other.

If instead you want to argue that \( n + 1 \not \mid n! \), then consider \( n = 5 \).

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Thanks sir, but am I in the right track to moving on or i should change my methods ? Thanks in advance !

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You are most likely on the wrong track, in that I believe there is no solution arising from your current approach.

See Pi Han's comment stream for suggestions on how to solve this problem.

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A simpler approach is to apply Bertrand's Postulate.

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That is an overkill for this problem. There is a much simpler approach.

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Yes. It involves the highest power of 2.

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