Induction Proof of Series SumDate: 02/03/2001 at 11:30:27 From: Jigar Subject: Induction How can I prove the following for all n >= 2? P(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n) >= 7/12 It is easy to prove the base condition n = 2, but I cannot find the right approach for the inductive step. I have never seen an induction proof where the right-hand side is a constant. Please help. Date: 02/03/2001 at 16:21:59 From: Doctor Anthony Subject: Re: Induction If n = 2 we have: 1/3 + 1/4 = 7/12 So result is true in the case n = 2. Assume that it is true for n = k. So we assume: P(k) = 1/(k+1) + 1/(k+2) + ... + 1/2k >= 7/12 Then if n = k+1 we have the series: P(k+1) = 1/(k+2) + 1/(k+3) + ... + 1/(2k) + 1/(2k+1) + 1/(2k+2) = [P(k) - 1/(k+1)] + 1/(2k+1) + 1/(2k+2) = P(k) + 1/(2k+1) - 1/(2k+2) = P(k) + [(2k+2)-(2k+1)]/[(2k+1)(2k+2)] = P(k) + 1/[(2k+1)(2k+2)] > 7/12 + 1/[(2k+1)(2k+2)] > 7/12 So if the result is true for n = k, then it is also true for n = k+1. But it is true for n = 2, and therefore it is true for n = 3, and if it is true for n = 3, then it is also true for n = 4 and so on for all positive integral values of n. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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