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)
This can be vizualized almost immediately once the inequality is written in the equivalent form
1/2+1/3+...+1/n < Ln(n) < 1+1/2+...+1/(n1)
since Ln(n) is the integral from 1 to n of dx/x and the sums on the
left and right respectively underestimate and overestimate the integral.
A fancy proof can be given based on integrating from k to k+1 the inequality
1/(k+1) < 1/x < 1/k
that holds for x in the open interval (k,k+1) to yield
1/(k+1) < integral from k to k+1 of dx/x < 1/k. Then summing from k=1 to n1 gives the desired result.
Edited on August 5, 2006, 3:11 am

Posted by Richard
on 20060803 21:04:08 