For polynomials $a=a(x)$ and $b=b(x)$, define the continued fraction $$f(a,b):=a(1)+ \lower 2pt\overset{\infty }{\underset{n=1}{\mathbb{\LARGE K}}}~\dfrac{b(n)}{a(n+1)}=a(1)+\cfrac{b(1)}{a(2) + \cfrac{b(2)}{a(3)+\lower 2pt\ddots } }.$$ Then it is well known and immediate from Euler's continued fraction formula that for $r>1$, we have $$\frac1{\zeta(r)}=f\Bigl(x^r+(x-1)^r,-x^{2r}\Bigr),$$ e.g. $$\frac1{\zeta(3)}=1-\cfrac{1^6}{1^3+2^3 - \cfrac{2^6}{2^3+3^3-\lower 2pt\ddots } }.$$ Now I have found numerically a generalization of this for $\zeta(3)$, more precisely, a family of continued fractions with slightly different polynomials $a(x)$ each time, and having as closed forms each "tail" of the $\zeta(3)$ series: for $k\in\mathbb N$, it looks in fact like $$ \frac1{\displaystyle{ \sum_{j=k}^{\infty}\frac1{j^3}}}=\frac1{\displaystyle{ \zeta(3)-\sum_{j=1}^{k-1}\frac1{j^3}}}= {f\Bigl(x^3+(x-1)^3+2k(k-1)(2x-1),-x^{6}\Bigr)}= {f\Bigl(\bigl[2x-1\bigr]\bigl[x(x-1)+k^2+(k-1)^2\bigr],-x^{6}\Bigr)}.$$ While for $k=1$ this reduces to the entire series $\zeta(3)$ as above, I wonder how to prove this for $k\ge 2$. And very intriguingly, in spite of the simple form (which, once discovered, looks quite straightforward), I haven't yet found if/how this can be generalized to other zeta values at integers, not even for $\zeta(2)$.
EDIT (23.07.22) By means of the polygamma identity $$\sum_{j=k}^{\infty}\frac1{j^3}=-\frac12\psi^{(2)}(k)$$ or equivalently just using a Hurwitz zeta function with integer shift $k\in\mathbb N$, it appears that the conjectured identity above can be extended to half-integers $k\in\mathbb Z+\frac12$. If we "blow up" the polynomials of the continued fraction to have integer coefficients again, we get, after shifting by $\frac12$ to $k\in\mathbb N$, the next conjecture for the tail of the "odd $\zeta(3)$ sum": $$ {4~\displaystyle{\sum_{j=k}^{\infty}\frac1{(2j-1)^3}}}= {4\left({\displaystyle{ \frac78\zeta(3)-\sum_{j=1}^{k-1}\frac1{(2j-1)^3}}}\right)}=\\=\frac1{f\Bigl(2x^3+2(x-1)^3+(4k^2-1)(2x-1),-4x^{6}\Bigr)}= \frac1{f\Bigl(\bigl[2x-1\bigr]\bigl[ x^2+(x-1)^2+4k^2\bigr],-4x^{6}\Bigr)} .$$
EDIT (22.06.22). There are very similar CFs not for $\zeta(2)$, but for the tail of the alternating sum $\eta(2)$, as mentioned in the comments (even this seems to be still unproven): $$\sum_{j=k}^{\infty}\frac{(-1)^{j+k}}{j^2}=\frac2{f\Bigl(x^2+(x-1)^2+k(k-1),-x^4\Bigr) } .$$
EDIT (23.07.22) Like for the $\zeta(3)$ sum above, this can also be extended to half-integers after separating even and odd terms and using the Hurwitz Zeta Function, where the sum on Wolfram's RHS (dropping the $(-1)^k$ of course) then can be written in terms of the Catalan constant $G$. It appears indeed that for $k\in\mathbb N$ (but surprisingly, not for $k=0$), it is simply the tail of the Catalan sum $\beta(2)$, which has conjecturally the following continued fraction representation: $$2\sum_{j=k}^{\infty}\frac{(-1)^{k+j}}{(2j+1)^2}=2\left|G-\sum_{j=0}^{k-1}\frac{(-1)^{j}}{(2j+1)^2}\right|=\frac1{f\Bigl(4x^2+4(x-1)^2+4k^2-1 , -16x^{4}\Bigr)}. $$
Note that the convergence rate of all these continued fractions is about as slow as for $f\Bigl(x^3+(x-1)^3,-x^{6}\Bigr)$.
