How can one demonstrate that $\lim_{n\to \infty} \sum_{k=1}^{n}{2\over 2k-1}-\ln{4n}=\gamma$ is correct? Let:

$$\lim_{n\to \infty} \sum_{k=1}^{n}{2\over 2k-1}-\ln{4n}=\gamma\tag1$$

Where $\gamma$ is Euler-Mascheroni Constant
How can one demonstrate that $(1)$ is correct?
Normally $\gamma$ it is defined  by $$\lim_{n\to  \infty}\sum_{k=1}^{n}{1\over k}-\ln{n}=\gamma \tag2$$
Numerically check-out that $(1)$ converges faster than $(2)$
 A: Denote the first sequence by $ a_n $ and the second sequence by $ b_n $, then we have that
$$ a_n - b_{2n} = \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} - \log(2) \to 0 $$
as $ n \to \infty $. Since $ b_{2n} \to \gamma $, it follows that $ a_n \to \gamma $ as well.
A: Once we know that
$$ H_n = \sum_{k=1}^{n}\frac{1}{k} = \log(n) + \gamma +\frac{1}{2n}-\frac{1}{12n^2}+O\left(\frac{1}{n^4}\right) \tag{A}$$
we also have that:
$$ \begin{eqnarray*}\sum_{k=1}^{n}\frac{2}{2k-1}&=& \sum_{k=1}^{2n}\frac{2}{k}-\sum_{k=1}^{n}\frac{2}{2k}\\&=&2 H_{2n}-H_n\\&=&\log(4n)+\gamma+\frac{1}{24n^2}+O\left(\frac{1}{n^4}\right)\end{eqnarray*}\tag{B}$$
with an improved convergence, due to the absence of the $\frac{1}{n}$ term.
We may further improve the speed of convergence, for instance by considering that
$$ 8H_{4n}-6H_{2n}+H_n-3\log(n) = 3\gamma + 10\log(2)+O\left(\frac{1}{n^4}\right)\tag{C} $$
from which:

$$ \color{red}{3\gamma} = \lim_{n\to +\infty}\left[-3\log(n)-10\log(2)+\sum_{k=1}^{n}\left(\frac{8}{4k-3}-\frac{4}{4k-2}+\frac{8}{4k-1}\right)\right]\tag{D} $$

$(D)$ is a sort of slow-converging BBP-formula for $\gamma$.
A: $$\sum_{k=1}^{n}{2\over 2k-1}$$
$$=2 \sum_{k=1,\text{odd}}^{2n-1} \frac{1}{k}$$
$$=2 \sum_{k=1}^{2n-1} \frac{1-(-1)^k}{2} \frac{1}{k}$$
$$=\sum_{k=1}^{2n-1} \frac{1}{k}+\sum_{k=1}^{2n-1} \frac{(-1)^{k+1}}{k}$$
Thus,
$$\sum_{k=1}^{n}{2\over 2k-1}-\ln (2n-1)-\ln (2+\frac{2}{2n-1})$$
\begin{align*}= \left(\sum_{k=1}^{2n-1} \frac{1}{k}-\ln (2n-1) \right)+\sum_{k=1}^{2n-1} \frac{(-1)^{k+1}}{k}-\ln (2+\frac{2}{2n-1}) \end{align*}
$$\to \gamma+\ln 2-\ln 2$$
$$=\gamma$$
