Let \begin{equation*} \Gamma(z)=\int_0^{\infty}t^{z-1}\textrm{e}^{-t}\textrm{d}t, \quad \Re(z)>0 \end{equation*} and $$ \psi(z)=[\ln\Gamma(z)]'=\frac{\Gamma'(z)}{\Gamma(z)}. $$ In the literature, these two functions are respectively called the Euler gamma function and the digamma function. The derivatives $\psi^{(k)}(z)$ for $k=1,2,3,4$ are respectively called the trigamma, tetragamma, pentagamma, and hexagamma functions. For non-specific positive integer $n\in\mathbb{N}$, we call $\psi^{(n)}(z)$ the polygamma function.

I have proved that, for $n\in\mathbb{N}$ and $\alpha>0$, the ratio $\frac{\psi^{(n)}(x+\alpha)}{\psi^{(n)}(x)}$ is increasing from $(0,\infty)$ onto $(0,1)$. This is equivalent to saying that, for $n\in\mathbb{N}$ and $\alpha>0$, the function $$ \Phi_{n,\alpha}(x)=\frac{1}{1-\frac{\psi^{(n)}(x+\alpha)}{\psi^{(n)}(x)}} $$ is increasing from $(0,\infty)$ onto $(1,\infty)$.

After considering the increasing property of $\Phi_{n,\alpha}(x)$, I would like to ask the following question.

For $n\in\mathbb{N}$ and $\alpha>0$, is the function $\Phi_{n,\alpha}(x)$ convex on $(0,\infty)$?

The answer to this question should be YES.

This problem is perhaps much easy for you, but it is very difficult for me right now. Please have a try to solve it. Thank you very mcuh.

I have asked this question at the site on the ResearchGate, but till now I haven't gotten any correct and useful answer.