From the series $\sum_{n=1}^{+\infty}\left(H_n-\ln n-\gamma-\frac1{2n}\right)$ to $\zeta(\frac12+it)$. Here is a pretty series 

$$
\displaystyle \sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)=\frac{1}{2} \left(1-\ln (2\pi)+\gamma\right) \tag{*}
$$

where $H_{n}:=\sum_{1}^{n} \frac{1}{k}$ are the harmonic numbers and $\gamma := \lim\limits_{n \to \infty} (H_n- \ln n)$ is the Euler constant.
$$ $$

Now just introduce a parameter in the general term of the series and you get a link with... the Riemann $\zeta$ function on the critical line!

Q 1. What proof would you give for (*)? 
Q 2. What elements would you give to get the link with $\zeta\left(\frac{1}{2}+it\right)$?
 A: Since
$$
1+\sum_{k=2}^n\left(\frac1k-\log\left(\frac{k}{k-1}\right)\right)=H_n-\log(n)
$$
we have
$$
\begin{align}
H_n-\log(n)-\gamma
&=\sum_{k=n+1}^\infty\left(\log\left(\frac{k}{k-1}\right)-\frac1k\right)\\
&=\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac1{k+1}\right)
\end{align}
$$
Furthermore,
$$
\frac1{2n}=\sum_{k=n}^\infty\frac12\left(\frac1k-\frac1{k+1}\right)
$$
Therefore,
$$
H_n-\log(n)-\gamma-\frac1{2n}=\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)
$$
Summing, we have
$$
\begin{align}
&\sum_{n=1}^\infty\left(H_n-\log(n)-\gamma-\frac1{2n}\right)\\
&=\sum_{n=1}^\infty\sum_{k=n}^\infty\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\
&=\sum_{k=1}^\infty\sum_{n=1}^k\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\
&=\sum_{k=1}^\infty k\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)
\end{align}
$$
and
$$
\begin{align}
&\sum_{k=1}^nk\left(\log\left(\frac{k+1}k\right)-\frac12\left(\frac1{k+1}+\frac1k\right)\right)\\
&=\sum_{k=2}^{n+1}(k-1)\log(k)-\sum_{k=1}^nk\log(k)-\sum_{k=1}^n\left(1-\frac1{2(k+1)}\right)\\
&=\color{#C00000}{n\log(n+1)}\color{#00A000}{-\log(n!)-n}\color{#0000F0}{+\frac12(H_{n+1}-1)}\\
&=\color{#C00000}{n\log(n)+1+O\!\left(\frac1n\right)}\color{#00A000}{-\left(n+\frac12\right)\log(n)-\frac12\log(2\pi)+O\!\left(\frac1n\right)}\\
&\color{#0000F0}{+\frac12\log(n)+\frac12\gamma-\frac12+O\!\left(\frac1n\right)}\\
&=\frac12\left(1+\gamma-\log(2\pi)\right)+O\!\left(\frac1n\right)
\end{align}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\left(H_n-\log(n)-\gamma-\frac1{2n}\right)=\frac{1+\gamma-\log(2\pi)}2}
$$
A: Proof of (*)
Adding the four finite sums,
$$\sum_{k=1}^{n}H_{k}=(n+1)H_{n}-n,$$
$$\sum_{k=1}^{n}\ln{k}=\ln{n!},$$
$$\sum_{k=1}^{n}\gamma=\gamma\,n,$$
$$\sum_{k=1}^{n}\frac{1}{2k}=\frac12H_{n},$$
gives us a representation of the $n$-th partial sum for the infinite series. Writing the infinite series as the limit of partial sums, we get:
$$\begin{align}
S
&=\sum_{k=1}^{\infty}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\
&=\lim_{n\to\infty}\sum_{k=1}^{n}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\
&=\lim_{n\to\infty}\left((n+1)H_{n}-n-\ln{n!}-\gamma\,n-\frac12H_{n}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right).
\end{align}$$
Use Stirling's approximation for the factorial to obtain an asymptotic formula for the log-factorial term in the series:
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\
\implies \ln{n!}\sim\ln{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)}=\left(n+\frac12\right)\ln{n}-n+\frac12\ln{(2\pi)}.$$
Then,
$$\begin{align}
S
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\left(n+\frac12\right)\ln{n}+n-\frac12\ln{(2\pi)}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-\gamma\,n-\left(n+\frac12\right)\ln{n}\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}\left(n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\left(H_{n}-\ln{n}\right)\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\lim_{n\to\infty}\left(H_{n}-\ln{n}\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\gamma-\frac12\ln{(2\pi)}\\
&=\frac12+\frac12\gamma-\frac12\ln{(2\pi)}.~~~\blacksquare
\end{align}$$

Appendix:
Using the asymptotic series for the digamma function given by Eq.16 on this Wolfram Mathworld page,
$$\begin{align}
\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)
&=\lim_{n\to\infty}n\left(\Psi{(n+1)}-\ln{n}\right)\\
&=\lim_{n\to\infty}n\left(\frac{1}{2n}-\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell}}\right)\\
&=\frac12-\lim_{n\to\infty}\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell-1}}\\
&=\frac12.
\end{align}$$
A: Observe that 
$$
H_{n}-\ln n-\gamma -\frac{1}{2n} = \psi (n) - \ln n + \frac{1}{2n}
$$
where $\psi := \Gamma'/\Gamma$ is the digamma function, using $\displaystyle \psi (n)=  H_{n-1}-\gamma = H_n-\gamma- \frac{1}{n}$, $n\geq 1$. 
Our initial series thus rewrites 
$$
\sum_{n=1}^{\infty} \left( \psi(n )- \log n + \frac{1}{2n}\right)  = \frac{\gamma}{2} - \frac{1}{2}\ln(2\pi)+ \frac{1}{2},
$$
(proved by David H).
Then consider the one parameter series
$$
\sum_{n=1}^{\infty}\left(\psi(n \alpha)- \log (n \alpha) + \frac{1}{2n \alpha}\right), \quad \alpha >0.
$$
We have the following result.

Theorem 1. Let $\alpha$ and $\beta$ be positive real numbers such that $ \alpha\beta=1$. 
Then
  \begin{align}
&\sqrt{\alpha}\left\{\frac{\gamma-\log(2\pi\alpha)}{2\alpha}+ \sum_{n=1}^{\infty}\left(\psi(n \alpha)- \log (n \alpha) + \frac{1}{2n \alpha}\right)\right\}\\
= & \sqrt{\beta}\left\{\frac{\gamma-\log(2\pi\beta)}{2\beta}+\sum_{n=1}^{\infty}\left(\psi(n \beta)- \log (n \beta) + \frac{1}{2n \beta}\right)\right\} \\
= &-\frac{1}{\pi^{3/2}}\int_0^{\infty}\left|\xi\left(\frac{1}{2}+\frac{it}{2}\right)\Gamma\left(\frac{-1+it}{4}\right)\right|^2
\frac{\cos\left(\frac{t}{2}\log\alpha\right)}{1+t^2}dt,
\end{align}

where  $$ \xi(s):=\frac{s(s-1)}{2} \displaystyle \pi^{-s/2}\:\Gamma(\frac{s}{2})\zeta(s)$$ and where $\zeta$ is the Riemann zeta function. 
Now express $\displaystyle  \left|\xi\left(\frac{1}{2}+\frac{it}{2}\right)\right|^2 $ in terms of $\left|\zeta \left(\frac{1}{2}+ it\right)\right|^2$ and you obtain the evocated link.
Theorem 1 is due to Ramanujan and one may find a recent proof here.
Here is a related result I have found.

Theorem 2. Let $\Re \alpha >0$.
Then
  $$ \sum_{n=1}^{\infty} \! \left(\! \psi(\alpha n )- \log (\alpha n ) + \frac{1}{2 \alpha n }\! \right)\! =\! \displaystyle \frac{1+\gamma-\log(2\pi)}{2} \\ -\int_{0}^{1}\left(\frac{1}{\alpha (1-x^{1/\alpha})}-\frac{1}{1-x}+\frac{1}{2}-\frac{1}{2\alpha}\!\right)\!\frac{\mathrm{d}x}{1-x}.$$

Thanks.
A: This is not an answer but contains information that may be useful in building one.
There are closed form expressions for $H_n-log\left(n\right)$ and $\gamma$ that come from generalized Mercator series.
$$
H_n-log\left(n\right)=1-\sum_{k=1}^{\infty}\left(\sum_{i=nk+1}^{n(k+1)}\frac{1}{i}-\frac{1}{k+1}\right)
$$
https://math.stackexchange.com/a/1602945/134791
$$
\gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)
$$
https://math.stackexchange.com/a/1591256/134791
