Practicing Success
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 $N=1,2,3, ....g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=$ |
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2 N-1)^2}\right\}$ $4\left\{1+\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2 N-1)^2}\right\}$ $-4\left\{1+\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2 N+1)^2}\right\}$ $4\left\{1+\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2 N+1)^2}\right\}$ |
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+...+\frac{1}{(2 N-1)^2}\right\}$ |
We have, $g(x)=\log f(x)$ and $f(x+1)=x f(x)$ $\Rightarrow g(x+1)=\log f(x+1)$ $\Rightarrow g(x+1)=\log (x f(x))$ $[∵ f(x+1)=x f(x)]$ $\Rightarrow g(x+1)=\log x+\log f(x)$ $\Rightarrow g(x+1)-g(x)=\log x$ $\Rightarrow g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^2}$ Putting $x=\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2}, ..., \frac{2 N-1}{2}$, we get $g^{\prime \prime}\left(1+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4$ $g^{\prime \prime}\left(2+\frac{1}{2}\right)-g^{\prime \prime}\left(1+\frac{1}{2}\right)=-\frac{4}{9}$ $g^{\prime \prime}\left(3+\frac{1}{2}\right)-g^{\prime \prime}\left(2+\frac{1}{2}\right)=-\frac{4}{25}$ $g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(N-\frac{1}{2}\right)=-\frac{4}{(2 N-1)^2}$ Adding all these, we get $g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left(1+\frac{1}{9}+\frac{1}{25}+\frac{1}{(2 N-1)^2}\right)$ |