From the expansion expressed in the OP, we have
$$NH_N-N\log\left(N+\frac12\right)-N\gamma =O\left(\frac1N\right) \tag1$$
We will exploit the expansion in $(1)$ in the following development.
We wish to evaluate the series $S$ as given by
$$S=\sum_{n=1}^\infty \left(H_n-\log\left(n+\frac12\right)-\gamma\right)$$
To proceed, we examine the terms in the sequence of partial sums $S_N$
$$S_N=\sum_{n=1}^N \left(H_n-\log\left(n+\frac12\right)-\gamma\right) \tag 2$$
The first term on the right-hand side of $(2)$ can be written as
$$\begin{align}
\sum_{n=1}^N H_n&=\sum_{n=1}^N \sum_{k=1}^n\frac1k\\\\
&=\sum_{k=1}^N \left(\frac{1}{k} \sum_{n=k}^{N}(1)\right)\\\\
&=\sum_{n=1}^N \frac{N-n+1}{n}\\\\
&=(N+1)H_N-N \tag 3
\end{align}$$
The second term on the right-hand side of $(2)$ can be written as
$$\begin{align}
\sum_{n=1}^N \log\left(n+\frac12\right)&=\log\left((2N+1)!!\right)-N\log(2)\\\\
&=\log\left(\frac{(2N+1)!}{2^N\,N!}\right)-N\log(2)\\\\
&=\log\left(\frac{\sqrt{2\pi(2N+1)}\left(\frac{2N+1}{e}\right)^{2N+1}}{\sqrt{2\pi N}\left(\frac{N}{e}\right)^N}\right)-2N\log(2)+O\left(\frac1N\right)\\\\
&=\frac12 \log\left(2+\frac1N\right)\\\\
&+(2N+1)\log\left(2N+1\right)-N\log(N)\\\\
&-(N+1)-2N\log(2)+O\left(\frac1N\right)\\\\
&=\frac12 \log(2)\\\\
&+(2N+1)\log(2)+(N+1)\log(N)+1\\\\
&-(N+1)-2N\log(2)+O\left(\frac1N\right)\\\\
&=\frac32\log(2)+(N+1)\log(N)-N+O\left(\frac1N\right)\tag 4
\end{align}$$
Using $(3)$ and $(4)$ in $(2)$ reveals
$$\begin{align}
S&=\lim_{N\to \infty}S_N\\\\
&=\color{blue}{\lim_{N\to\infty}N\left(H_N-\log\left(N+\frac12\right)-\gamma\right)}\\\\
&+\color{red}{\lim_{N\to\infty}(H_N-\log(N))}\\\\
&+\color{green}{\lim_{N\to\infty}N\left(\log\left(N+\frac12\right)-\log(N)\right)}\\\\
&-\frac32\log(2)+\lim_{N\to\infty}\left(O\left(\frac1N\right)\right)\\\\
&=\color{blue}{0}+\color{red}{\gamma}+\color{green}{\frac12}+0\\\\
&=\gamma +\frac{1-3\log(2)}{2}
\end{align}$$
as was to be shown!
