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.
 A: Observe that, by absolute convergence:
$$\begin{align}
\sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)& =\sum_{k=1}^{\infty} \dfrac{1+(-1)^k}{2}\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right)\\\\
&=\frac{1}{2}\sum_{k=1}^{\infty} \!\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\!\right)\right)\!+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right) \\\\
&=\frac{\gamma}{2}+\!\frac{1}{2}\!\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right).
\end{align}
$$
Then, as $N$ is great, write
$$
\begin{align}
\sum_{k=1}^{N} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)& =\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\sum_{k=1}^{N} (-1)^{k-1}\log\!\left(1+\dfrac{1}{k}\!\right)\\\\
&=\sum_{k=1}^{N}(-1)^{k-1} \frac{1}{k}-\log \left(\prod_{k=1}^{N} \left(1+\dfrac{1}{k}\!\right)^{(-1)^{k-1}}\right) \tag2
\end{align}
$$ giving
$$
\begin{align}
\sum_{k=1}^{\infty} (-1)^{k-1}\left(\!\frac{1}{k} -\log\!\left(1+\dfrac{1}{k}\!\right)\!\right)=\log 2-\log \left(\frac{\pi}{2}\right)
\end{align}
$$
where we have used Wallis' product formula for $\pi$.
Finally we obtain

$$
\sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)=\frac{\gamma}{2}-\frac{1}{2}\log \left(\frac{4}{\pi}\right).
$$

A: 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}
A: 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
$$\sum_{k=1}^\infty \left[\frac{1}{2k} - \log\left(1+\frac{1}{2k}\right) \right]
= \frac12\left(\gamma + \log\pi\right) - \log 2
$$
A: If I have well understood, you want the limit of $\displaystyle S_n=\sum_{k=1}^n (\frac{1}{2k}-\log(1+\frac{1}{2k}))$.
We have 
$$\sum_{k=1}^n \frac{1}{k}=\log(n+1)+\gamma+\varepsilon_n$$ with $\varepsilon_n\to 0$ as $n\to +\infty$.
Now:
$$S_n=\frac{1}{2}(\sum_{k=1}^n\frac{1}{k})-\sum_{k=1}^n \log(2k+1)+\sum_{k=1}^n \log(2k)=\frac{1}{2}(\sum_{k=1}^n\frac{1}{k})-\sum_{k=1}^n (\log(2k+1)+\log(2k))+2\sum_{k=1}^n \log(2k)$$
Hence:
$$S_n=\frac{\gamma}{2}+\frac{1}{2}\log(n+1)+\frac{\varepsilon_n}{2}-\log(2n+1)!+2n\log 2+2\log n!$$
Thus:
$$S_n=\frac{\gamma}{2}+\frac{\varepsilon_n}{2}+\log A_n$$
With $\displaystyle A_n=\frac{2^{2n}(n!)^2\sqrt{n+1}}{(2n+1)!}$.
Now use Stirling's Formula to finish. 
A: Notice that $$S=\sum^{\infty}_{k=1}\Big(\frac{1}{2k}-\frac{1}{2}\log(1+\frac{1}{k})+\frac{1}{2}\log(1+\frac{1}{k})-\frac{2}{2}\log(1+\frac{1}{2k})\Big)$$
Therefore 
$$S=\frac{1}{2}\gamma+\frac{1}{2}\sum^{\infty}_{k=1}\log\Big(\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)$$
In other words
\begin{align}
S=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\prod^{\infty}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{1+\frac{1}{k}}{(1+\frac{1}{2k})^2}\Big)\\&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\prod^{N}_{k=1}\frac{k+1}{k}\frac{(2k)^2}{(2k+1)^2}\Big)\\
&=\frac{1}{2}\gamma+\frac{1}{2}\log\Big(\lim_{N\to\infty}\frac{\pi\Gamma(N+1)\Gamma(N+2)}{4\Gamma^2(N+3/2)}\Big)\\
&=\frac{1}{2}\gamma+\frac{1}{2}\log(\frac{\pi}{4})
\end{align}
We have used the limiting value $$\lim_{N\to\infty}\frac{\Gamma(N+1)\Gamma(N+2)}{\Gamma^2(N+3/2)}=1$$
