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Induction Proof of Series Sum


Date: 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/   
    
Associated Topics:
High School Number Theory

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