Question

If $s_n=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{n},\left(nϵN\right)$, then $s_1+s_2+s_3+s_4+...s_n=(n+\lambda )s_{n+1}-(n+1)$. Find the value of $\lambda$

Answer 1

$P=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}$
$=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}asb=\frac{2ac}{a+c}$
$=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}(\frac{c}{a}+\frac{a}{c})\ge 4$
So, $\sqrt{\lambda \sqrt{\lambda \sqrt{\lambda ........\infty }}}={\lambda }^{\frac{1}{2}+\frac{1}{4}+\frac{1}{b}}+.....\infty ={\lambda }^{\frac{1/2}{1-1/2}}=\lambda$
$\Rightarrow \lambda =4$