How do we findS=∑k=1∞[12k−log⁡(1+12k)]

Consider the series
${\displaystyle S=\sum _{k=1}^{\mathrm{\infty }}[\frac{1}{2k}-\ln (1+\frac{1}{2k})]}$
for which, by using the logarithm in series form, it becomes
${\displaystyle S=\sum _{k=1}^{\mathrm{\infty }}[\frac{1}{2k}-\frac{1}{2k}+\sum _{n=2}^{\mathrm{\infty }}\frac{{(-1)}^n}{n{\left(2k\right)}^n}]}$
${\displaystyle =\sum _{k=1}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{{(-\frac{1}{2})}^n}{n{k}^n}}$
${\displaystyle =\sum _{n=2}^{\mathrm{\infty }}\frac{{(-\frac{1}{2})}^n}{n}\zeta \left(n\right)}$
${\displaystyle =\sum _{n=1}^{\mathrm{\infty }}\frac{{(-\frac{1}{2})}^{n+1}}{n+1}\zeta (n+1)}$
Now the generating function for the zeta function is given by
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\zeta (n+1){(-1)}^{n+1}{x}^n=\gamma +\psi (x+1)}$
and by integration leads to
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{{(-x)}^{n+1}}{n+1}\zeta (n+1)=\gamma x+\ln \mathrm{\Gamma }(x+1)}$
Letting ${\displaystyle x=\frac{1}{2}}$ in this series leads to the value of the series for S, namely
${\displaystyle \sum _{k=1}^{\mathrm{\infty }}[\frac{1}{2k}-\ln (1+\frac{1}{2k})]=\frac{1}{2}\ln \left(\frac{\pi }{4}\right)+\frac{\gamma }{2}}$