Proof Involving Sums of Reciprocals

Date: 09/13/2004 at 11:29:14
From: Drew
Subject: Sums of Reciprocals

For n > 1, prove that the sum of reciprocals from 1 to n does not 
sum to an integer.  For n = 4, for example, prove that 1/1 + 1/2 + 
1/3 + 1/4 does not sum to an integer.

I have tried a few methods of approach.  I have tried the most 
obvious, induction.  I found a theorem that says that if gcd(a, b) = 
gcd(c, d) = 1, then a/b + c/d is an integer only if the absolute 
value of b and d is equal.  So, in the inductive step, I am adding 
on 1/(k+1) which fits the c and d above, but I can't think how to 
prove that when you reduce the "k" step (from the inductive 
hypothesis), that the denominator of that fraction is not equal to 
(k+1).  That was my induction work.  

I have tried some other things too.  Some numerical analysis to see 
if there was some pattern to the fractions, but there seems to be 
none.

Date: 09/13/2004 at 14:18:42
From: Doctor Vogler
Subject: Re: Sums of Reciprocals

Hi Drew,

Thanks for writing to Dr. Math.  I would do this by starting with

   n   1
  sum --- = M
  i=1  i

and then multiplying both sides by n!/k (where k divides n!) so that
everything is an integer, including all of the terms of the sum on the
left *except one*.  Bertrand's Postulate can be helpful here, but you
don't even need that.  You can do it with k being just the right power
of two.

Does that help?

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/