Evaluate Difference of Second Derivatives of Logarithmic Gamma Related Function

Given that
$g(x) = \log f(x)$
where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ satisfying the functional relation:
$f(x + 1) = x f(x)$

Taking the natural logarithm on both sides of the relation, we get:
$\log f(x + 1) = \log(x \cdot f(x))$
$\implies \log f(x + 1) = \log x + \log f(x)$

Substituting $g(x) = \log f(x)$ into the equation gives:
$g(x + 1) = \log x + g(x)$

Differentiating both sides with respect to $x$:
$g'(x + 1) = \frac{1}{x} + g'(x)$

Differentiating again with respect to $x$:
$g''(x + 1) = -\frac{1}{x^2} + g''(x)$
$\implies g''(x + 1) - g''(x) = -\frac{1}{x^2}$

To evaluate $g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right)$, we substitute $x = k + \frac{1}{2}$ for $k = 0, 1, 2, \dots, N-1$:

$g''\left(k + \frac{3}{2}\right) - g''\left(k + \frac{1}{2}\right) = -\frac{1}{\left(k + \frac{1}{2}\right)^2} = -\frac{4}{(2k + 1)^2}$

Summing both sides from $k = 0$ to $N - 1$:
$\sum_{k=0}^{N-1} \left[ g''\left(k + \frac{3}{2}\right) - g''\left(k + \frac{1}{2}\right) \right] = \sum_{k=0}^{N-1} \left[ -\frac{4}{(2k + 1)^2} \right]$

The left-hand side forms a telescoping series:
$\left[g''\left(\frac{3}{2}\right) - g''\left(\frac{1}{2}\right)\right] + \left[g''\left(\frac{5}{2}\right) - g''\left(\frac{3}{2}\right)\right] + \dots + \left[g''\left(N + \frac{1}{2}\right) - g''\left(N - \frac{1}{2}\right)\right]$
$= g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right)$

Evaluating the right-hand side sum:
$-4 \sum_{k=0}^{N-1} \frac{1}{(2k + 1)^2} = -4\left\{1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N - 1)^2}\right\}$

Thus,
$g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right) = -4\left\{1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{(2N - 1)^2}\right\}$

Hence, the correct option is (A).