let Z(n)=Ln(n)(1/2 + ... + 1/n)
Prove that 0 < Z(n) < 11/n, where n is an integer greater then one (Ln(n) being naturalbase logarithm of n)
I will be using the following property that was found at this website
http://mathworld.wolfram.com/HarmonicNumber.htmlit 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)<1y(1/[2(n+1)])
and
(3) Z(n)>[2n(1y)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 20060803 20:18:03 