# Is my solution correct?

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 : $1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} (\forall n \in \mathbb{Z})$ Note by Rony Phong
4 years, 5 months ago

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Good 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 $n = 5$.

Staff - 4 years, 5 months ago

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

- 4 years, 5 months ago

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.

Staff - 4 years, 5 months ago

A simpler approach is to apply Bertrand's Postulate.

- 4 years, 5 months ago

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

Staff - 4 years, 5 months ago

Yes. It involves the highest power of 2.

- 4 years, 5 months ago