# 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
3 years, 8 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 - 3 years, 8 months ago

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

- 3 years, 8 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 - 3 years, 8 months ago

A simpler approach is to apply Bertrand's Postulate.

- 3 years, 8 months ago

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

Staff - 3 years, 8 months ago

Yes. It involves the highest power of 2.

- 3 years, 8 months ago