Strange result about the log sum I am working in a infinity sum and I get the strange result
$$\sum _{n=1}^{\infty } \frac{1}{2} \log \left(\frac{1}{n^2}\right)=\log (2 \pi )$$
it seem as $$-2 \zeta '(0)$$ but i do not justify?
it maybe wrong it came from
$$\frac{1}{2} \log \left(\frac{1}{n^2}\right)+\log (x)-\log (2)-\log (\pi )=-\log (x)$$
 A: Simplified view:
Consider the more general series
\begin{align}
S_{p} = \sum_{n=1}^{\infty} \frac{1}{2 \, n^{p}} \, \ln\left(\frac{1}{n^{2}}\right).
\end{align}
By using $\ln\left(\frac{1}{n^{2}}\right) = - \ln(n^{2}) = - 2 \, \ln(n)$ then
\begin{align}
S_{p} &= - \, \sum_{n=1}^{\infty} \frac{\ln(n)}{n^{p}} 
= \sum_{n=1}^{\infty} \partial_{p} \left(\frac{ 1}{n^{p}}\right) 
= \partial_{p} \left(\zeta(p)\right)
\end{align}
By letting $p \to 0$ then this leads to
$$S_{0} = \zeta^{\prime}(0) = - \frac{1}{2} \, \ln(2 \pi)$$.  

Another view:
\begin{align}
S_{0} &= \frac{1}{2} \, \sum_{n=1}^{\infty} \ln\left(\frac{1}{n^{2}}\right) = - \sum_{n=1}^{\infty} \ln(n) = - \lim_{p \to \infty} \ln\left[ \prod_{n=1}^{p} \{n\} \right] = - \lim_{p \to \infty} \left( \ln(p!) \right)
\end{align}
From here a number of approximations can be utilized to evaluate the limit. 

For more information:


*

*$\zeta^{\prime}(0)$ is equation (38) in Zeta Function 

*A Wallis Product derivation of $\zeta^{\prime}(0)$ is given in Wallis Product
