Question

Să se arate că lg(1-1/2)+lg(1-1/3)+lg(1-1/4)+...+lg(1-1/1000)<0

Answer 1

$\lg (1-\frac{1}{2} ) +\lg (1-\frac{1}{3} )+\lg (1-\frac{1}{4})+\ldots +\lg (1-\frac{1}{1000})<0\\\lg (\frac{1}{2} )+\lg (\frac{2}{3})+\lg (\frac{3}{4} )+\ldots +\lg (\frac{999}{1000})<0\\\lg (\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}\cdot ....\cdot \frac{999}{1000})<0\\\lg (\frac{1}{1000})<0\\\\-\lg (1000)<0\\-3<0 (A)$

Answer 2

lg(1 -1/2) +lg(1 -1/3) +lg(1 -1/4) +...+lg(1 -1/1000)=㏒₁₀(1/2) +㏒₁₀(2/3) +㏒₁₀(3/4) +...+㏒₁₀        (999/1000) =㏒₁₀(1/2 ·2/3 ·3/4 ·...·999/1000)=㏒₁₀(1/1000)=㏒₁₀10⁻³=-3 < 0 .