1@kumar anirudha
please post all such things in olympiad section
the below given solution was given by my 9th grader friend
Bertrand's Postulate: \exists p\in\mathbb P \forall n\in\mathbb N satisfying n<p<2n
We can equivalently say that \left\lfloor \frac n2 \right\rfloor <p <n.
Now assume
S_n =\sum_{i=1}^n \frac 1 i\in\mathbb N.\\ We \ can \ say \ that \ \\ S_n =\frac 1{n!}\sum_{i=1}^n \frac{n!}{i}.
From the Bertrand's Postulate we must have a prime p that lies between \left\lfloor \frac n2\right\rfloor and n; which satisfies p | n! (follows directly from our assumption
If so, according to our claim that Sn is integer, we must have
p | n! \left| \sum_{i=1}^n \frac{n!}{i}.
So we see that all the terms of the form \frac{n!}{i} is divisible by p except p.
Therefore we must have p\left| \frac{n!}{p}; which forces n\geq 2p
We can have .
In both cases, we see that n \geq2p leads to a contradiction.
Hence proved
after all this , i think i have to go back to class 8 [2]
