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        riddles >> medium >> Another Inequality
        
(Message started by: ThudanBlunder on Jan 14th, 2009, 7:39am)
Title: Another Inequality
Post by ThudanBlunder on Jan 14th, 2009, 7:39am
For n > 1 prove that 1 + 1/22 + 1/32 + 1/42 + ....... + 1/n2  >  3n/(2n + 1)
Title: Re: Another Inequality
Post by towr on Jan 14th, 2009, 8:36am
[hide]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.[/hide]
Title: Re: Another Inequality
Post by 1337b4k4 on Jan 14th, 2009, 11:22am
[hide] 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)[/hide]


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