TITLE: Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$
QUESTION [7 upvotes]: 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.

REPLY [20 votes]: Start with the infinite product expansion of Gamma function,
$$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}}$$
Taking logarithm and rearrange terms, one get
$$\begin{align}
& -\log\Gamma(z) = \log z + \gamma z + \sum_{k=1}^\infty \left[\log\left(1+\frac{z}{k}\right) - \frac{z}{k}\right]\\
\implies & \sum_{k=1}^\infty \left[\frac{z}{k} - \log\left(1+\frac{z}{k}\right) \right] = \log z + \gamma z + \log\Gamma(z)
\end{align}$$
Taking $z = \frac12$ and use the known value of $\Gamma\left(\frac12\right) = \sqrt{\pi}$, we get
$$\sum_{k=1}^\infty \left[\frac{1}{2k} - \log\left(1+\frac{1}{2k}\right) \right]
= \frac12\left(\gamma + \log\pi\right) - \log 2
$$