Maximum of a one-parameter function family. It is well known that determining maxima or minima of a function is a difficult task even if the function depends on one variable only. Having said that we consider a following function:
\begin{equation}
\mu(\alpha) := \sqrt{\alpha(2-\alpha)} \cdot \left( \Psi(1-\alpha,-2 H,2) - 2  \Psi(1-\alpha,-2 H,1) +  \Psi(1-\alpha,-2 H,0)\right) \frac{1}{2}
\end{equation}
where $\Psi(z,s,a) := \sum\limits_{k=0}^\infty z^k (k+a)^{-s}$ is the Hurwitz-Lerch transcendent.  Below we show a contour-plot of the function in question.
 
Clearly for fixed $H$ our function increases monotonically as a square root for small values of $\alpha$ and decreases for bigger values of $\alpha$. As such it has only one unique maximum. Interestingly enough this maximum seems to depend linearly on the parameter $H$. In fact we have verified numerically that that maximum reads:
\begin{equation}
\alpha_{max} = \left\{ 
\begin{array}{rr}
1.52746 - 2.12184 \cdot H & \mbox{for $0.5 < H < 0.75$}\\
0 & \mbox{$H \ge 0.75$}
\end{array}
\right.
\end{equation} 
This linear relationship is quite puzzling because it holds not only in some vicinity of $H=1/2$ but over the whole range $H \in (0.5,0.72)$ and it deviates from it only for $0.72 < H \le 0.75$. The question is therefore how do we explain this linear relationship. It is just a coincidence or is it due to some deeper properties of the Hurwitz-Lerch transcendent?
 A: Before we give the answer we will simplify the definition of our function a little bit. Even a short glimpse at the expression in brackets amounts to conclude that that expression is somehow related to a discrete second derivative. By replacing that discreet derivative by a continuous one we can substantially simplify our expression.  Clearly we have:
\begin{equation}
\Delta_k^2 (k+1)^{2 H} \simeq H (2 H-1) \cdot (k+1)^{2H-2}
\end{equation}
where $\Delta_k$ is an operator of a discreet derivative and $k=0,1,2,\cdots$. By taking both this and the definition of the Lerch function we can approximate the right hand side as:
\begin{equation}
\mu(\alpha) = H (2 H-1)\cdot \sqrt{\alpha(2-\alpha)} \cdot \Psi(1-\alpha,2-2 H,1)
\end{equation}
We will be using the form from now on. We differentiate the above equation with respect to $\alpha$ and we set it to zero. We have:
\begin{equation}
F[\alpha,H] := \frac{1}{H (2 H-1)}\frac{d \mu(\alpha)}{d \alpha} =  \frac{(\alpha-2) \alpha \text{Li}_{1-2 H}(1-\alpha)+\text{Li}_{2-2 H}(1-\alpha)}{(\alpha-1)^2 \sqrt{(2-\alpha) \alpha}} = 0
\end{equation} 
The equation above defines an implicit function $\alpha(H)$ being the location of the maxima. Firstly we take the limit $H \rightarrow 1/2$. This leads to a simple transcendental equation $-\log(\alpha) = (2-\alpha)(1-\alpha)$ which has a solution $\alpha=\alpha_{1/2} = 0.31619737624797931884580421116257156770718671236239...$.
Now we expand our implicit function about $H=1/2$. We have:
\begin{eqnarray}
\left.\frac{d \alpha(H)}{d H} \right|_{H=1/2} &=& \left.-\frac{\partial_H F}{\partial_a F}\right|_{H=1/2,\alpha=\alpha_{1/2}} = -1.29645988377449161733..\\
\left.\frac{d^2 \alpha(H)}{d H^2} \right|_{H=1/2} &=& \left. -\frac{\partial_H F\cdot [\partial_a F]^2-2 \partial^2_{H,a}F \cdot\partial_H  F \cdot \partial_a F+ \partial_a F \cdot [\partial_H F]^2}{\partial_a^3F} \right|_{H=1/2,a=a_{1/2}} = -2.478365541856920625560..\\
\left.\frac{d^3 \alpha(H)}{d H^3} \right|_{H=1/2} &=& \cdots =2.44826460914193254661108..
\end{eqnarray}
In the last equation we have dropped the explicit expression for the derivative since it is unwieldy and does not fit into this page yet of course it is easily coded in Mathematica for example.
Below we plot the maxima location. Here Blue, Red and Maroon correspond to a first order, second order and third order expansion respectively.

As we can see the linear term provides a good approximation of the underlying function yet it overshoots at big values of $H$.  The whole series has not yet converged at the third order and inclusion of higher terms is necessary. The overall conclusion is that the function in question is not a linear function and it is just a coincidence that it looks linear.
