Dear all,


I have been trying to analytically prove a statement (which should hold) for a couple of days now, but am no longer making any progress. Any insights would therefore be very much appreciated. The statement is the following:


Given that $\displaystyle $g\left( x,k\right) =\frac{x^{k}}{1-\alpha }$$, where $\displaystyle $\alpha =\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}$$, and $\displaystyle $x\geq 2$$ and $\displaystyle $k\geq 1$$, show that $\displaystyle $g\left( x,k+1\right) >g\left( x,k\right) $$.

My progress

Spoiler:


This is equivalent to showing $\displaystyle $g\left( x,k+1\right) -g\left( x,k\right) >0$,$ or

$\displaystyle $x^{k+1}\frac{1}{1-\frac{Ln\left( x^{k+2}-1\right) }{x^{k+2}-2}}-x^{k}\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}>0$$

$\displaystyle $\Longleftrightarrow x^{k}\left( \frac{x}{1-\frac{Ln\left( x^{k+2}-1\right)}{x^{k+2}-2}}-\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}\right) >0$$

Since $\displaystyle $x^{k}>0$$, it suffices to show that $\displaystyle $\left( \frac{x}{1-\frac{Ln\left( x^{k+2}-1\right)}{x^{k+2}-2}}>\frac{1}{1-\frac{Ln\left( x^{k+1}-1\right) }{x^{k+1}-2}}\right) $$

$\displaystyle $\Longleftrightarrow\left( \frac{1}{x}\right) \left( \frac{Ln\left(x^{k+2}-1\right) }{x^{k+2}-2}-1\right) >\frac{Ln\left( x^{k+1}-1\right)}{x^{k+1}-2}-1$$

Now I take out the logarithms:


$\displaystyle $\Longleftrightarrow Ln\left( \left( x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }\right)-\frac{1}{x}>Ln\left( \left( x^{k+1}-1\right) ^{\left( \frac{1}{^{x^{k+1}-2}}\right) }\right)-1$$

$\displaystyle $\Longleftrightarrow\frac{x-1}{x}>Ln\left( \frac{\left( x^{k+1}-1\right)^{\left( \frac{1}{^{x^{k+1}-2}}\right) }}{\left( x^{k+2}-1\right)^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}\right) $$

$\displaystyle $\Longleftrightarrow Exp\left( \frac{x-1}{x}\right) >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$$

Now it suffices to show that:


$\displaystyle $Exp\left( \frac{x-1}{x}\right) >\frac{3}{2} >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$$

First part:


$\displaystyle $Exp\left( \frac{x-1}{x}\right) >\frac{3}{2} $$ if $\displaystyle x>1.67$


Second part (here's where I am stuck):


$\displaystyle \frac{3}{2} >\left( x^{k+1}-1\right) ^{\left( \frac{1}{{x^{k+1}-2}}\right) }\frac{1}{\left(x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }}$$