Proving $\zeta(2) - \beta(1) + \zeta(4) - \beta(3) + \zeta(6)- \beta(5) + \ldots=1$

Trying to prove

$$\zeta(2) - \beta(1) + \zeta(4) - \beta(3) + \zeta(6)- \beta(5) + \ldots=1$$ I found by numerical calculation that (when $k$ goes to infinity)

$$\sum_{n=1}^{k}\zeta (2n)=k+3/4+o(1),$$

$$\sum_{n=1}^{k}\beta (2n-1)=(k-1)+3/4+o(1).$$

in order to prove the above formula. Now how can I prove it analytically?

We have: $$\zeta(2k) = \frac{1}{(2k-1)!}\int_{0}^{+\infty}\frac{t^{2k-1}}{e^t-1} \tag{1} $$ and: $$\beta(2k-1)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2k-1}}=\frac{1}{(2k-2)!}\int_{0}^{+\infty}\frac{t^{2k-2}e^t}{e^{2t}+1}\,dt \tag{2}$$ hence: $$\sum_{n\geq 1} \beta(2n-1)\, z^{2n-1} = \frac{\pi z}{4}\cdot\sec\frac{\pi z}{2},\tag{3} $$ $$\sum_{n\geq 1} \zeta(2n)\,z^{2n} = \frac{1-\pi z \cot(\pi z)}{2},\tag{4} $$ and: $$ \lim_{z\to 1^-}\sum_{n\geq 1}\left(\zeta(2n)\, z^{2n}-\beta(2n-1)\, z^{2n-1}\right) = \color{red}{1}.\tag{5}$$