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Possible Duplicate:
Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer?

Does there exist a proof, not by induction, of the fact that $1 + (1/2) + \cdots + (1/n)$ is not an integer for any $n > 1$? In all the books I have read the proof is done by induction and is not quite illuminating . Without induction the most I have been able to get is ;

$(d(1) + d(2) + \cdots + d(n) ) /n ≤ 1 + (1/2) + \cdots + (1/n) < 1+ (d(1) + d(2) + \cdots + d(n) )/n$; where $d(k)$ denotes the number of positive divisors of $k$ , can this inequality be improved or, adequately altered to prove the theorem ? Does there exist a proof of the theorem which does not use Bertrand's postulate neither order properties nor induction ?

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marked as duplicate by JavaMan, The Chaz 2.0, Rudy the Reindeer, Paul, Marc van Leeuwen Oct 12 '12 at 7:09

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  • $\begingroup$ See this page if it helps en.wikipedia.org/wiki/Harmonic_number $\endgroup$ – Hawk Oct 12 '12 at 4:52
  • $\begingroup$ @jak:- The link is not helping! $\endgroup$ – Souvik Dey Oct 12 '12 at 5:04
  • $\begingroup$ @ JavaMan:- Don't think my post as a duplicate , I would like to know if the theorem can be proved by improving or, altering the inequality I stated in my post and if it's not possible I would like to know if there is a proof without using Bertrand's postulate neither induction nor the order properties ( as shown in the proof of the post you linked) $\endgroup$ – Souvik Dey Oct 12 '12 at 5:14
  • $\begingroup$ Your post literally says "Does there exist a proof , not by induction , of the fact that..." As such, the topic has already been discussed and is therefore an exact duplicate. Your post never mentioned Bertrand's postulate or "order properties." $\endgroup$ – JavaMan Oct 12 '12 at 6:44
  • $\begingroup$ @ JavaMan:- Well,I think I have fixed my question suitably now. $\endgroup$ – Souvik Dey Oct 12 '12 at 7:03

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