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Topic: Another inequality (Read 839 times) |
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anonymous
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Let a1 > 3 be a real number.
Define an+1 = (an)^2 - nan + 1 for n=1,2,3,...
Prove that sum(n=1 to n=infinity) 1/( 1 + an ) < 1/2
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towr
wu::riddles Moderator
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Re: Another inequality
« Reply #1 on: Jun 24th, 2003, 2:05pm » |
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1/( 1 + an ) <= 1/2n+1 for all n, so 1/2 * sum(1/2i, i, 1, inf) =1/2 is the upper limit for the sum
to prove it, I need to prove that
1 + an >= 2^(n+1) for all n
1 + a1 >= 4 is a given since an >= 3
so
an+1 +1 >= 2 + 2*an >= 2^(n+2)
an2 -n*an +2 >= 2 + 2*an
an -n >= 2
an >= 2 + n
needs to be true
which is easily proven by
a1 >= 2 + 1 and
an+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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