

Title: Another inequality Post by anonymous on Jun 24^{th}, 2003, 12:09pm Let a_{1} > 3 be a real number. Define a_{n+1} = (a_{n})^2  na_{n} + 1 for n=1,2,3,... Prove that sum(n=1 to n=infinity) 1/( 1 + a_{n} ) < 1/2 

Title: Re: Another inequality Post by towr on Jun 24^{th}, 2003, 2:05pm 1/( 1 + a_{n} ) <= 1/2^{n+1} for all n, so 1/2 * sum(1/2^{i}, i, 1, inf) =1/2 is the upper limit for the sum to prove it, I need to prove that 1 + a_{n} >= 2^(n+1) for all n 1 + a_{1} >= 4 is a given since a_{n} >= 3 so a_{n+1} +1 >= 2 + 2*a_{n} >= 2^(n+2) a_{n}^{2} n*a_{n} +2 >= 2 + 2*a_{n} a_{n} n >= 2 a_{n} >= 2 + n needs to be true which is easily proven by a_{1} >= 2 + 1 and a_{n+1} >= (2+n)^{2}  n(2+n) + 1 = 2n +5 > 2 + n from there it's a small step from 'sum <= 1/2' to 'sum < 1/2' 

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