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 Inequality (Posted on 2006-08-03)
let Z(n)=Ln(n)-(1/2 + ... + 1/n)

Prove that 0 < Z(n) < 1-1/n, where n is an integer greater then one (Ln(n) being naturalbase logarithm of n)

 No Solution Yet Submitted by atheron No Rating

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 my solution | Comment 1 of 3
I will be using the following property that was found at this website
http://mathworld.wolfram.com/HarmonicNumber.html
it is equation 14.

This property states that
(1) 1/[2(n+1)]<H(n)-Ln(n)-y<1/(2n)

where H(n) is the nth harmonic number or in other words
H(n)=1/1+1/2+1/3+....1/n

and where y is eulers constant which for the purpose of this proof I will use the aproximation of y=0.577

now first I will make some usefull definitions

Z(n)=Ln(n)-(1/2+...1/n)

I will set H'(n)=(1/2+....1/n)  now it is easy to see that
H(n)=1+H'(n)

now using this to substitute into (1) doing some simple rearranging of the inequalities we have

(2) Z(n)<1-y-(1/[2(n+1)])
and
(3) Z(n)>[2n(1-y)-1]/2n

now it is simple to show that if n>=2 we have the righthand side of (2) less than 1-(1/n) and the righthand side of (3) greater than 0 and thus we have

Z(n)<1-(1/n) and Z(n)>0 and thus we have the desired inequality

0<Z(n)<1-(1/n)

 Posted by Daniel on 2006-08-03 20:18:03

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