Let $g(x) = \log f(x)$ where $f(x)$ is twice differentiable positive function on $(0, \infty)$ such that $f(x+1) = x f(x)$. Then, for $N = 1, 2, 3, \ldots$ $g''\left(N + \frac{1}{2}\right) - g''\left(\frac{1}{2}\right) =$

$g(x) = \log f(x),\quad f(x) > 0,\quad f(x+1) = x f(x)$ $x \in (0, \infty),\qquad N = 1, 2, 3, \ldots$ $\text{To find: } g''\!\left(N + \frac{1}{2}\right) - g''\!\left(\frac{1}{2}\right)$ $\text{Plan: }g(x+1) = g(x) + \log x$ $g''(x+1) - g''(x) = -\dfrac{1}{x^{2}}$ $\text{telescope from } \tfrac{1}{2} \text{ to } N+\tfrac{1}{2}$ $g(x+1) = \log\bigl(x f(x)\bigr) = \log x + \log f(x) = \log x + g(x)$ $g(x+1) - g(x) = \log x$ $g'(x+1) = g'(x) + \dfrac{1}{x}$ $g''(x+1) = g''(x) - \dfrac{1}{x^{2}}$ $g''(x+1) - g''(x) = -\dfrac{1}{x^{2}}$ $g''\!\left(N+\tfrac{1}{2}\right) - g''\!\left(\tfrac{1}{2}\right) = \sum_{k=1}^{N}\left[g''\!\left(k+\tfrac{1}{2}\right) - g''\!\left(k-\tfrac{1}{2}\right)\right]$ $g''\!\left(k+\tfrac{1}{2}\right) - g''\!\left(k-\tfrac{1}{2}\right) = -\dfrac{1}{\left(k-\frac{1}{2}\right)^{2}} = -\dfrac{4}{(2k-1)^{2}}$ $\sum_{k=1}^{N} -\dfrac{4}{(2k-1)^{2}} = -4\left\{1 + \dfrac{1}{9} + \dfrac{1}{25} + \cdots + \dfrac{1}{(2N-1)^{2}}\right\}$ $\boxed{\text{option (1)}}$