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Inequality (Posted on 2006-08-03) Difficulty: 1 of 5
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)

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Solution another solution Comment 3 of 3 |

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 n-1 and summing the inequalities we obtain


The first inequality is equivalent with 0<Z(n). The second is equivalent with Ln(n)-(1/2+...+1/(n-1))<1. Substracting 1/n in both members it results Z(n)<1-1/n

  Posted by Stefan on 2006-08-17 18:22:43
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