Proposed:
$$\gamma=\lim_{n \to \infty}\left\{\left[\,\sum_{k=1}^{n}{\zeta\left(\,2k\,\right)\over k}\,\right] -\ln\left(\,2n\,\right)\right\}\tag1$$
Where $\gamma$ is Euler-Mascheroni Constant
My try
Well-known $$\gamma=\lim_{n\to \infty}\left(\sum_{k=1}^{n}{1\over k}\right)-\ln(n)\tag2$$
How do we prove $(1)?$
On
Using the given formula (2) it is sufficient to prove that $$\sum_{k=1}^{\infty} \frac{1}{k} (\zeta(2k) - 1 ) = \ln 2.$$ So now set $f(x) := \sum_{k=1} ^{\infty} \frac{1}{k} (\zeta(2k) - 1 ) x^{2k}$. Then $$xf'(x) = 2 \sum_{k=1}^{\infty} (\zeta(2k) - 1 ) x^{2k} = - \pi x \cot(\pi x ) + 1 - \frac{2x^2}{1 - x^2}.$$ Therefore, integrating $f'(x)$, $$f(1-\epsilon) - f(\epsilon) = \ln(1-\epsilon) - \ln(1 -\epsilon^2) +\ln(2 - \epsilon)$$ for $0<\epsilon<1$. Letting $\epsilon \to 0$ we get the desired conclusion.
Note that $$\sum_{k\geq1}\frac{\zeta\left(2k\right)-1}{k}=\sum_{k\geq1}\frac{1}{k}\sum_{m\geq2}\frac{1}{m^{2k}}$$ $$=\sum_{m\geq2}\sum_{k\geq1}\frac{1}{km^{2k}}=-\sum_{m\geq2}\log\left(1-\frac{1}{m^{2}}\right)$$ $$=-\log\left(\prod_{m\geq2}\left(1-\frac{1}{m^{2}}\right)\right)$$ and it is well known that $$\prod_{m\geq2}\left(1-\frac{1}{m^{2}}\right)=\frac{1}{2}$$ since $$\prod_{m=2}^{N}\left(1-\frac{1}{m^{2}}\right)=\prod_{m=2}^{N}\frac{m^{2}-1}{m^{2}}=\prod_{m=2}^{N}\frac{m-1}{m}\prod_{m=2}^{N}\frac{m+1}{m}=\frac{1}{N}\frac{N+1}{2}$$ so $$\sum_{k\geq1}\frac{\zeta\left(2k\right)-1}{k}=\color{red}{\log\left(2\right)}.$$ The result obviously implies your limit since $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{n}\frac{\zeta\left(2k\right)}{k}-\log\left(2n\right)\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{n}\frac{\zeta\left(2k\right)}{k}-\sum_{k=1}^{n}\frac{1}{k}+\sum_{k=1}^{n}\frac{1}{k}-\log\left(2n\right)\right)$$ $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{n}\frac{\zeta\left(2k\right)-1}{k}+\sum_{k=1}^{n}\frac{1}{k}-\log\left(n\right)-\log\left(2\right)\right)=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^{n}\frac{1}{k}-\log\left(n\right)\right)=\color{blue}{\gamma}.$$