You can do it by calulating the n-th partial sum and taking the limit:
Let $S_n=\sum_{k=1}^{n} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$. Then:
$$
S_n=\frac{1}{2}H_n -\log\left(\prod_{k=1}^n\dfrac{2k+1}{2k}\right)=\frac{1}{2}H_n -\log\left(\dfrac{(2n+1)!!}{(2n)!!}\right)=\frac{1}{2}H_n -\log\left(\dfrac{(2n+1)!}{2^{2n}(n!)^2}\right)
$$
By Stirlings approximation, we have
$$
n!=\sqrt{2\pi}\cdot n^{n+\frac{1}{2}}\cdot e^{-n}\cdot e^{\frac{\theta_n}{12n}}
$$
With $0\lt\theta_n\lt1$. Making this substitution in the fomula above, we get:
$$
S_n=\frac{1}{2}H_n -\log\left(\dfrac{\sqrt{2\pi}\cdot(2n+1)^{2n+1+\frac{1}{2}}\cdot e^{-2n-1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{2^{2n}\cdot 2\pi\cdot n^{2n+1} \cdot e^{-2n}\cdot e^{\frac{\theta_n}{6n}}}\right)=\frac{1}{2}H_n -\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot \sqrt{n+\frac{1}{2}}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right)=\frac{1}{2}H_n -\frac{1}{2}\log\left(n+\frac{1}{2}\right)-\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right)
$$
So by taking the limit, we get:
$$
S=\lim_{n \to \infty}S_n=\lim_{n \to \infty}{\left[\frac{1}{2}H_n -\frac{1}{2}\log\left(n+\frac{1}{2}\right)-\log\left(\dfrac{2\cdot(1+\frac{1}{2n})^{2n+1}\cdot e^{\frac{\theta_{2n+1}}{12(2n+1)}}}{\sqrt{\pi}\cdot e\cdot e^{\frac{\theta_n}{6n}}}\right)\right]=\frac{\gamma}{2}-\log(2)+\frac{1}{2}\log{(\pi)}}
$$
Which seems to be true since Wolfram alpha gives:
$$
\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right] \approx 0.1678255948155212079577375992595540032692269400673623
$$
While:
$$
\frac{\gamma}{2}-\log(2)+\frac{1}{2}\log{(\pi)} \approx 0.1678255948155212079577375992595540032692269400673623
$$