Given that,

$$g(x) = \log f(x)$$

$$ \Rightarrow g(x + 1) = \log f(x + 1)$$

$$ \Rightarrow g(x + 1) = \log xf(x)$$

[$$\because$$ $$f(x + 1) = xf(x)$$]

$$g(x + 1) = \log x + \log f(n)$$

$$ \Rightarrow g(x + 1) - g(x) = \log (x)$$

$$g'(x + 1) - g'(x) = {1 \over x}$$

$$ \Rightarrow g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$$

Putting, $$x = x - {1 \over 2}$$, we get

$$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) = - {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$$

Putting $$x = 1,2,3$$ .........., N we get

$$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) - = {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$$

Putting $$x = 1,2,3$$, .........., N we get

$$g''\left( {{3 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$$ .... (i)

$$g''\left( {{5 \over 2}} \right) - g''\left( {{3 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$$ ..... (ii)

$$g''\left( {{7 \over 2}} \right) - g''\left( {{5 \over 2}} \right) = {{ - {2^2}} \over {{5^2}}}$$ ..... (iii)

$$g''\left( {N + {1 \over 2}} \right) - g''\left( {N - {1 \over 2}} \right) = {{ - {2^2}} \over {{{(2N - 1)}^2}}}$$ ..... (iv)

Adding all the above equations, we get

$$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = 4$$

$$\left[ {1 + {1 \over {{3^2}}} + {1 \over {{5^2}}}\, + \,...\, + \,{1 \over {{{(2N - 1)}^2}}}} \right]$$