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Topic: Another Inequality (Read 415 times) |
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ThudnBlunder
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Another Inequality
« on: Jan 14th, 2009, 7:39am » |
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For n > 1 prove that 1 + 1/22 + 1/32 + 1/42 + ....... + 1/n2 > 3n/(2n + 1)
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towr
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Re: Another Inequality
« Reply #1 on: Jan 14th, 2009, 8:36am » |
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We can simply use induction.
3(n+1)/[2(n+1) + 1] - 3n/(2n + 1) = 3 / (4n2 + 8n +3 )
1/(n+1)2 > 3 / (4n^2 + 8n +3 ) for n > 0
Therefore, if sum1..n 1/i2 > 3n/(2n + 1) then sum1..n+1 1/i2 > 3(n+1)/[2(n+1) + 1]
It is true for n=2, so it must be true for all n.
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| « Last Edit: Jan 14th, 2009, 8:43am by towr » |
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teekyman
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Re: Another Inequality
« Reply #2 on: Jan 14th, 2009, 11:22am » |
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you could also hand check for the first 7 values, after which
1 + 1/4 + 1/9 + ... > 1 + 1/4 + 1/9 + ... 1/49 > 1.5 > 3n/(2n+1)
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| « Last Edit: Jan 14th, 2009, 11:22am by teekyman » |
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