I am very happy to find the "tails" formula of $\zeta(3)$ here. I have discovered the "tails" formula of $\eta(2)$ and $\beta(2)$ (Catalan's constant). You should be able to find these two similar formula from the identities below:
$$
\zeta(2)=2\eta(2)=1+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{4}{1+\frac{6}{1+\frac{9}{1+\frac{12}{1+\cdots }}}}}}}
$$
$$
\beta(2)=\frac{1}{2}+\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{\frac{1}{2}+\frac{2}{\frac{1}{2}+\frac{4}{\frac{1}{2}+\frac{6}{\frac{1}{2}+\frac{9}{\frac{1}{2}+\frac{12}{\frac{1}{2}+\cdots}}}}}}}
$$

OEIS A087811 $\left \{1,2,4,6,9,12,16,20,25,30,36,42,49,56,64,72,⋯\right \}$

$a_{k}=k+a_{k-2},  a_0=1,  a_1=2$

I'd like to know if there is a "tails" formula for $\eta(3)$.