On the monotonicity of the function $z \mapsto (z^{1-s}-1)/(z^s-1)$.

Question

During my work I have met the function $$ z \mapsto \frac{z^{1-s}-1}{z^s-1}, $$ which I consider for $z>1$ (say). The number $s \in (0,1)$ is a given parameter. I would like to prove that it is monotonically decreasing when $1/2 <s<1$, as some numerical investigation seems to confirm. I have tried to compute the first derivative, but at least three terms appear with competing signs, and I am unable to show that it is negative. Any help is appreciated.

Answer 1

Write

$$f(z)=\frac{z^{1-s}-1}{z^s-1} =\frac{z-z^s}{z^{2s}-z^s} $$

Now, for dixed $z$ and given $\epsilon>0$, observe that $f(z)<f(z+\epsilon)$ if and only if

$$\underbrace{z^{2s}(z+\epsilon)^{}+z^{}(z+\epsilon)^{s}+z^{s}(z+\epsilon)^{2s}}_{g(\epsilon)}< \underbrace{z^{2s}(z+\epsilon)^{s}+z^{s}(z+\epsilon)^{}+z^{}(z+\epsilon)^{2s}}_{h(\epsilon)}$$

Observe that $g(0)=h(0)=z^{1+2s}+z^{1+s}+z^{3s}$. On the other hand, we have that

$$g'(\epsilon)=z^{2s}+zs(z+\epsilon)^{s-1}+z^s2s(z+\epsilon)^{2s-1}$$ $$h'(\epsilon)=z^{2s}s(z+\epsilon)^{s-1}+z^s+z2s(z+\epsilon)^{2s-1}$$

So that

$$g'(0)=z^{2s}+sz^s+2sz^{3s-1} =z^s\left(s+z^s+2sz^{2s-1}\right)$$ $$h'(0)=sz^{3s-1}+z^s+2sz^{2s} =z^s\left(1+sz^{2s-1}+2sz^{s}\right)$$

I would like to show that $g'(0)<h'(0)$. This is true if and only if

\begin{align} s+z^s+2sz^{2s-1}&<1+sz^{2s-1}+2sz^{s}\\\iff sz+z^{1+s}+2sz^{2s}&<z+sz^{2s}+2sz^{1+s}\\\iff \underbrace{sz^{2s}}_{u(z)}&<\underbrace{(1-s)z+(2s-1)z^{1+s}}_{v(z)} \end{align}

Now, we have that $u(1)=v(1)=s$. On the other hand:

\begin{align} u'(z)&=2s^2z^{2s-1}\\ v'(z)&=(1-s)+(2s-1)(1+s)z^{s}\\ \end{align}

It is enough to show that $u'(z)<v'(z)$ for all $z>1$.

Observe that $u'(1)=v'(1)=2s^2$. It suffices hence to show that $u''(z)<v''(z)$ for all $z>1$. We have

\begin{align} u''(z)&=2s^2(2s-1)z^{2(s-1)}\\ v''(z)&=s(2s-1)(1+s)z^{s-1}\\ \end{align}

Now, $u''(z)<v''(z)\iff 2sz^{2(s-1)}<(1+s)z^{s-1}$. This is true, because

$$2s<1+s$$ $$z^{2(s-1)}<z^{s-1},$$

both of which are implied by $z>1$ and $s<1$. $\square$

Answer 2

Hint:  let $f(z)$ be the given function, then $\;\;\displaystyle f'(z) = \frac{s z^{2 s-1} - (2s-1) z^{s} + s - 1}{z^s(z^s - 1)^2}\,$. The denominator is positive, so the sign of $\,f'(z)\,$ is given by $\,g(z) = s z^{2 s-1} - (2s-1) z^{s} + s - 1\,$.

But $\,g(1) = 0\,$ and $\,g'(z)=s(2s-1)z^{s-1}(z^{s-1}-1) \lt 0\,$ for $\,z \gt 1, s \in (0,1)\,$, so $g(z) \lt 0$ for $z \gt 1\,$, and therefore $\,f'(z) \lt 0\,$ on $(1,\infty)$.