Question

Sa se demonstreze inegalitatea: 1/n+1 + 1/n+2 +.....+1/3n+1>1 AJUTOR URGENT

Answer 1

$S(n):\dfrac{1}{n+1} + \dfrac{1}{n+2}+...+\dfrac{1}{3n+1} > 1$

$P(1): \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}> 1\Rightarrow \dfrac{26}{24}>1\,\,\,(A)$

$P(n):\dfrac{1}{n+1} + \dfrac{1}{n+2}+...+\dfrac{1}{3n+1} > 1$

$P(n+1):\dfrac{1}{(n+1)+1} + \dfrac{1}{(n+1)+2}+...+\dfrac{1}{3(n+1)+1} > 1$

$\Rightarrow \dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+1}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}>1$

$\Rightarrow \underset{P(n)}{\underbrace{\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+1}}}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}>1+\dfrac{1}{n+1}$$\text{Dar:}$

$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+1}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}>\\ >1+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}$

$\Rightarrow 1+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4} > 1+\dfrac{1}{n+1}$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4} >\dfrac{1}{n+1}$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{3(n+1)}-\dfrac{1}{n+1}+\dfrac{1}{3n+4} >0$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{n+1}\cdot \left(1-\dfrac{1}{3}\right)+\dfrac{1}{3n+4} >0$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{n+1}\cdot \left(-\dfrac{2}{3}\right)+\dfrac{1}{3n+4} >0$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{3n+4} >\dfrac{2}{3(n+1)}$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{3n+4} >\dfrac{2}{3n+3}$

$\Rightarrow \dfrac{1}{3n+2}+\dfrac{1}{3n+4} >\dfrac{1}{3n+3}+\dfrac{1}{3n+3}$

$\Rightarrow \dfrac{6n+6}{(3n+2)(3n+4)}>\dfrac{6n+6}{(3n+3)^2}$

$\Rightarrow (3n+2)(3n+4) < (3n+3)^2$

$\Rightarrow 9n^2+12n+6n+8 < 9n^2+2\cdot 9\cdot n+9$

$\Rightarrow 9n^2+18n+8 < 9n^2+18n+9$

$\Rightarrow 8 < 9\,\,\,(A)$

(Ceea ce era de demonstrat)