Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is the Euler–Mascheroni constant.
But I could not manipulate this series.

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