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)
Applying the Lagrange's theorem for the function f(x)=Ln(x) on the interval [k,k+1], where k>=1, we obtain
Ln(k+1)Ln(k)=1/c, where k<c<k+1. But 1/(k+1)<1/c<1/k, so 1/(k+1)<Ln(k+1)Ln(k)<1/k.
Giving to k the values from 1 to n1 and summing the inequalities we obtain
1/2+...+1/n<Ln(n)<1+1/2+...+1/(n1)
The first inequality is equivalent with 0<Z(n). The second is equivalent with Ln(n)(1/2+...+1/(n1))<1. Substracting 1/n in both members it results Z(n)<11/n

Posted by Stefan
on 20060817 18:22:43 