The notation of $\psi$ means QPolyGamma symbol.
I noticed an interesting thing: zeros that computed by Mathematica
It seems that it has a lot of zeros; And I tried to plot the function out:
From the plot, we can deduce that only on the right side (or the left side) there would be zeros.
Next, I calculated the derivative of this function: $\displaystyle -\frac{\psi _2^{(2)}(n+2)}{\log 4}$
However, the derivative of the second one is $\displaystyle -\frac{\psi _2^{(3)}(n+2)}{\log 4}$. And finally this is not always $>0$, so $\displaystyle -\frac{\psi _2^{(2)}(n+2)}{\log 4}$ may have infinite sign changes.
This seems far from proving my original question, thus I didn't find any useful references to my question.
So that does $\displaystyle \frac{1}{2}-\frac{\psi _2^{(1)}(n+2)}{\log 4}$ have infinite zeros?
Write the $q$-polygamma function as a sum: $$\Gamma_2(z) = 2^{(z - 1) (z - 2)/2} \hspace{1px}\Gamma_{1/2}(z), \\ \psi_2(z) = \frac 1 {\Gamma_2(z)} \frac d {dz} \Gamma_2(z) = \psi_{1/2}(z) + z \ln 2 - \frac {3 \ln 2} 2 = -\ln 2 \left( \frac 1 2 - z + \sum_{k \geq 0} \frac 1 {2^{k + z} - 1} \right), \\ \frac 1 2 - \frac 1 {2 \ln 2} \frac d {dz} \psi_2(z) = -\frac {\ln 2} 2 \sum_{k \geq 0} \frac {2^{k + z}} {(2^{k + z} - 1)^2}.$$ Setting $z = n + 2$ gives your function. It's singular at $n = -2, -3, \dots$ and negative everywhere else.