Proof Involving Sums of ReciprocalsDate: 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/ |
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