How find this interesting sum $\sum_{n=1}^\infty\ln(f(n))+\ln(g(n))$ How  find this following interesting sum

$$\sum\limits_{n = 1}^\infty  {\left\{ {{{\ln }^2}\left( {n - \frac{1}{4}} \right) + {{\ln }^2}\left( {n{\rm{ + }}\frac{1}{4}} \right) - {{\ln }^2}\left( {n - \frac{1}{2}} \right) - {{\ln }^2}\left( {n + \frac{1}{2}} \right)} \right\}}  = \ln 2\ln \pi  - \frac{5}{2}{\ln ^2}2$$

I think we can use 
$$n!\approx\left(\dfrac{n}{e}\right)^n\sqrt{2n\pi}$$
and before I have solve following this
$$\sum_{n=1}^{\infty}\left(n\ln{(\dfrac{2n+1}{2n-1})}-1\right)=\dfrac{1-\ln{2}}{2}$$
 A: Here is my solution to the problem. The idea behind it is not difficult. It's the same thing that one can do to evaluate series like
$$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = 6$$
using generating functions. Note that 
$$\frac{d}{ds} \frac{1}{(n+a)^s} = \ln^2(n+a) \frac{1}{(n+a)^s}$$
We can view the series in question as the second derivative of function evaluated at 0.
So we're going to consider the following function
$$F(s) = \sum_{n=1}^{\infty} \frac{1}{(n+1/4)^s}+\frac{1}{(n-1/4)^s}-\frac{1}{(n+1/2)^s}-\frac{1}{(n-1/2)^s}$$
For $s$ with real part greater than 1 we can split the sum into a sum of Hurwitz zeta functions but we're not going to actually consider them explicitly since we'll calculate what the series are directly. Let's take care of each one.
$$\sum_{n=1}^{\infty}  \frac{1}{(n+1/4)^s} = 4^s \sum_{n=1}^{\infty} \frac{1}{(4n+1)^s} = 4^s\left(\sum_{n=0}^{\infty} \frac{1}{(4n+1)^s}-1\right) \tag{1}$$
$$\sum_{n=1}^{\infty}  \frac{1}{(n-1/4)^s} = 4^s \sum_{n=0}^{\infty} \frac{1}{(4n+3)^s} \tag{2}$$
$$\sum_{n=1}^{\infty}  \frac{1}{(n+1/2)^s} = 2^s \sum_{n=1}^{\infty}  \frac{1}{(2n+1)^s} = 2^s \left(\sum_{n=0}^{\infty}  \frac{1}{(2n+1)^s} - 1\right) \tag{3}$$
$$\sum_{n=1}^{\infty}  \frac{1}{(n-1/2)^s} = 2^s \sum_{n=1}^{\infty}  \frac{1}{(2n-1)^s} = 2^s \sum_{n=0}^{\infty}  \frac{1}{(2n+1)^s} \tag{4}$$
We combine $(1)$ and $(2)$
$$\sum_{n=1}^{\infty}  \frac{1}{(n+1/4)^s} + \sum_{n=1}^{\infty}  \frac{1}{(n-1/4)^s} = -4^s + 4^s\left(\sum_{n=0}^{\infty} \frac{1}{(4n+1)^s}+\sum_{n=0}^{\infty} \frac{1}{(4n+3)^s}\right) \\= -4^s + 4^s \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$
We combine this with $(3)$ and $(4)$ to get
$$F(s) = 2^s-4^s + (4^s-2\cdot2^s)\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$
The last series is not difficult to find a closed form of. We have
$$\zeta(s) = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} + \sum_{n=0}^{\infty}\frac{1}{(2n)^s} = \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} + 2^{-s}\zeta(s)$$
so
$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} = (1-2^{-s})\zeta(s)$$
Finally,
$$F(s)  =2^s-4^s + (4^s - 2\cdot 2^s)(1-2^{-s})\zeta(s) = (2^s-1)((2^s-2)\zeta(s)-2^s)$$
Using
$$\zeta(0) = -\frac{1}{2}$$
$$\zeta'(0) = -\frac{1}{2}\ln 2$$
we find that
$$F''(0) = \ln^2 2(-\zeta(0)-1)+2\ln 2(-\ln2+\ln2 \zeta(0)-\zeta'(0)) \\= -\frac{1}{2}\ln^2 2 + 2\ln 2\left(-\frac{3}{2}\ln 2 + \frac{1}{2}\ln(2\pi)\right)\\= \ln2 \ln \pi - \frac{5}{2}\ln^22$$
and we're done.
Note: We are actually doing things that are not explicitly allowed. For example, we're implicitly assuming that we can interchange the infinite summation with term by term differentiation and some other complex analytic phenomena. Someone with better analytic knowledge than me can hopefully point out justification as to why we're allowed to do all that.
Note 2: I'm thinking that we're really using something equivalent to Abel's theorem but with Dirichlet generating functions. Perhaps one could view this as using Perron's formula to convert to an ordinary generating function and then using Abel's theorem.
