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

Since, \(\quad f(x)=e^{g(x)}\)
\[
\begin{aligned}
\Rightarrow \quad e^{g(x+1)} & =f(x+1) \\
& =x f(x) \\
& =x e^{g(x)}
\end{aligned}
\]
and
\[
g(x+1)=\log x+g(x)
\]
\[
\Rightarrow \quad g(x+1)-g(x)=\log x
\]
Replacing \(x\) by \(x-\frac{1}{2}\), we get
\[
\begin{aligned}
g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\
\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =-\frac{4}{(2 x-1)^2}
\end{aligned}
\]
Substituting, \(x=1,2,3, \ldots, N\) in Eq. (ii) and adding, we get
\[
g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
\]