# Quotient of completely monotone functions

I am trying to show that a nontrivial quotient is (monotonically) decreasing in both of its arguments. More specifically:

$f\left( x,k\right) =\left( x^{k+1}-1\right) ^{\left( \frac{1}{x^{k+1}-2}\right) }$

$g\left( x,k\right) =\left( x^{k+2}-1\right) ^{\left( \frac{1}{x}\right) \left( \frac{1}{x^{k+2}-2}\right) }$

$x\geq2$

$k\geq0$

--> Show that f(x,k)/g(x,k) is decreasing in x and k

Since the derivatives of the quotient are not straightforward to calculate, I tried showing this indirectly by looking at f(x,k) and g(x,k) separately. More specifically, I was able to show that both f and g are completely monotonic in x and f. However, a quotient of completely monotonic functions is not necessarily (monotonically) decreasing. What kind of additional conditions would I require for the quotient to be decreasing? Is there a better way of showing the quotient is decreasing?

I would greatly appreciate any help.

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Welcome to math.stackexchange Abe! Some small advice I can give off the bat is to consider $\log(f/g) = \log f - \log g$ rather than $f/g$ directly. Since $\log$ is a monotonically increasing function, $f/g$ is decreasing if and only if $\log (f/g)$ is decreasing. The $\log$ brings down the exponents, which may make some things simpler. –  Ragib Zaman May 23 '12 at 12:02
thank you for this suggestion. I actually did it to other way around: I started with the logs, but could not prove that the difference in logs was decreasing so I turned it into a quotient. The logs are again completely monotonic, but their difference is not. –  Abe May 30 '12 at 16:00