How to prove that $\frac{\zeta(2) }{2}+\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}+\frac{\zeta (8)}{2^7}+\cdots=1$?

$$ \begin{align} \sum_{n=1}^\infty\frac{\zeta(2n)}{2^{2n-1}} &=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac2{k^{2n}2^{2n}}\tag{1}\\ &=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1{(4k^2)^n}\tag{2}\\ &=2\sum_{k=1}^\infty\frac1{4k^2-1}\tag{3}\\ &=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k+1}\right)\tag{4}\\[6pt] &=1\tag{5} \end{align} $$ Explanation:
$(1)$: expand $\zeta(2n)=\sum\limits_{k=1}^\infty\frac1{k^{2n}}$
$(2)$: change the order of summation
$(3)$: sum of a geometric series
$(4)$: partial fractions
$(5)$: telescoping sum

Since

$$\zeta(2n) = \frac{1}{(2n-1)!}\int_{0}^{\infty}\frac{x^{2n-1}}{e^x-1}\,dx $$

we have:

$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{2n-1}} = \int_{0}^{\infty}\frac{\sinh(x/2)}{e^x-1}\,dx =\frac12\int_{0}^{\infty}e^{-x/2}\,dx = \color{red}{1}.$$

The Laurent expansion of $\cot (z)$ at the origin in terms of the Riemann zeta function is $$ \cot (z) = \frac{1}{z} - 2 \sum_{k=1}^{\infty}\zeta(2k) \frac{z^{2k-1}}{\pi^{2k}} \ , \ 0 < |z| < \pi. $$

Letting $ \displaystyle z= \frac{\pi}{2}$, $$\cot \left(\frac{\pi}{2} \right) = \frac{2}{\pi} - \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{\zeta(2k)}{2^{2k-1}}.$$

But $\cot \left(\frac{\pi}{2} \right)=0$.

Therefore,

$$ \sum_{k=1}^{\infty} \frac{\zeta(2k)}{2^{2k-1}} = 1.$$