Show that
$ f(x) = \frac{\zeta(x -2)}{\zeta(x-1)} \qquad x > 3, $
where $\zeta$ is the Riemann-Zeta function, is non-increasing.
My attempt was to use $\zeta(s) = \frac{1}{\Gamma(s)} \int_0^{\infty} \frac{t^{s-1}}{e^{t}-1} \,dt$
Then
$f(x) = \frac{\Gamma(x-1)}{\Gamma(x-2)} \int_0^{\infty} \frac{t^{s-3}}{e^{t}-1} \,dt / \int_0^{\infty} \frac{t^{s-2}}{e^{t}-1} \,dt= (x-2)\int_0^{\infty} \frac{t^{s-3}}{e^{t}-1} \,dt / \int_0^{\infty} \frac{t^{s-2}}{e^{t}-1} \,dt$.
Any ideas on how to continue, or other suggestions? Thanks.
First of all note that$$\underset{n\geq1}{\sum}\frac{1}{n^{x-2}}\geq\underset{n\geq1}{\sum}\frac{1}{n^{x-1}}>0.\,\,\,\,(1)$$ We know that if $\Re\left(s\right)>1$ we have $$\zeta'\left(s\right)=-\underset{n\geq1}{\sum}\frac{\log\left(n\right)}{n^{s}}$$ so$$f'\left(x\right)=\frac{\zeta'\left(x-2\right)\zeta\left(x-1\right)-\zeta\left(x-2\right)\zeta'\left(x-1\right)}{\zeta\left(x-1\right)^{2}}\leq\frac{\zeta\left(x-2\right)\left(-\underset{n\geq1}{\sum}\frac{\log\left(n\right)}{n^{x-2}}+\underset{n\geq1}{\sum}\frac{\log\left(n\right)}{n^{x-1}}\right)}{\zeta\left(x-1\right)^{2}}\leq0$$ for the same reason of $(1).$