$\zeta(2)$
The inequality
$$9<\pi^2<10$$
can be obtained from series
$$\zeta(2)=\frac{\pi^2}{6}=\frac{3}{2}+\frac{1}{2}\sum_{k=0}^\infty \frac{1}{(k+1)^2(k+2)^2}\tag{1}$$
and
$$\zeta(2)=\frac{\pi^2}{6}=\frac{5}{3}-\sum_{k=1}^\infty \frac{1}{(k+1)(k+2)^2(k+3)}\tag{2}$$
Both of them have constant numerator and the denominator a polynomial of fourth degree.
$\zeta(3)$
Similarly, for Apéry's constant we can write the inequality
$$\frac{6}{5}<\zeta(3)<\frac{5}{4}$$
from series
$$\zeta(3)=\frac{6}{5}+\sum_{k=2}^\infty \frac{1}{k^3+4k^7}\tag{3}$$
and
$$\zeta(3)=\frac{5}{4}-\sum_{k=0}^\infty \frac{1}{(k+1)(k+2)^3(k+3)}\tag{4}$$
However, the degree of the polynomials is not the same in this case.
Is there a series for $\zeta(3)-\dfrac{6}{5}$ having the denominator a polynomial of order 5?
[EDIT]
From Jack D'Aurizio's study,
$$\begin{align}
\zeta(3)&=\frac{26}{21}-\frac{32}{7}\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
&=\frac{1138}{945}-\frac{32}{7}\sum_{k=1}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
&=\frac{198862}{165375}-\frac{32}{7}\sum_{k=2}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
&=\frac{9741838}{8103375}-\frac{32}{7}\sum_{k=3}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
&=\frac{2893129886}{2406702375}-\frac{32}{7}\sum_{k=4}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
&=\frac{4550782178678}{3785742835875}-\frac{32}{7}\sum_{k=5}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\
\end{align}$$
After James Arathoon's answer,
$$\begin{align}
\zeta(3)&=1+\frac{1}{2}\sum_{k=0}^\infty \frac{2k+3}{(k+1)^3(k+2)^3}\\
\\
&=1+\frac{1}{2}\sum_{k=1}^\infty \frac{2k+1}{k^3(k+1)^3}\\
\\
\zeta(3)&=\frac{6}{7}+\frac{64}{7}\sum_{k=0}^\infty \frac{k+1}{(2k+1)^3(2k+3)^3}\\
\\
&=\frac{6}{7}+\frac{64}{7}\sum_{k=1}^\infty \frac{k}{(4k^2-1)^3}\\
\end{align}$$
$\zeta(3)\approx \frac{6}{5}$ is a consequence of the continued fraction in formula (35), Apéry's Constant from Wolfram MathWorld.
Best Answer
I bet this can be considered as "cheating", but for any $a>0$ we have $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+a+1)^3(n+a+2)} = \frac{1+a}{2a^3}+\frac{1}{2a^3(a+1)}+\frac{1}{2}\psi''(a)\tag{1} $$ by partial fraction decomposition, and $\psi''(1)=\zeta(3), \psi''\left(\frac{1}{2}\right)=-14\,\zeta(3)$, $\psi''\left(\frac{3}{2}\right)=16-14\,\zeta(3), \psi''\left(\frac{5}{2}\right)=16+\frac{16}{27}-14\,\zeta(3)$ and
$$ \psi''\left(m+\frac{1}{2}\right)=16\left[\sum_{k=1}^{m}\frac{1}{(2k-1)^3}-\frac{7}{8}\zeta(3)\right]\tag{2} $$ By choosing a rather large value of $a$ like $a=\frac{7}{2}$, by $(1)$ and $(2)$ we get $$ 7\zeta(3) = \frac{9741838}{1157625}-\sum_{n\geq 0}\frac{1}{\left(n+\frac{7}{2}\right)\left(n+\frac{9}{2}\right)^3\left(n+\frac{11}{2}\right)}\tag{3} $$ where the involved series has a positive value, but less than $10^{-3}$, leading to an accurate approximation for $\zeta(3)$ of the wanted type. Similarly $$2\cdot 10^{-4}\approx\sum_{n\geq 0}\frac{1}{\left(n+\frac{11}{2}\right)\left(n+\frac{13}{2}\right)^3\left(n+\frac{15}{2}\right)}=\frac{4550782178678}{540820405125}-7\zeta(3)\tag{4}$$ and so on. By considering the continued fraction of $\frac{4550782178678}{3785742835875}$ and truncating it at the fourth convergent, we get $\zeta(3)\approx\color{red}{\frac{113}{94}}$ with an approximation error that is less than $10^{-4}$. $\frac{6}{5}$ is just the approximation we get by stopping at the third convergent.
It is interesting to point out that $\zeta(3)\approx\frac{6}{5}$ arises from the computation of the average order of the arithmetic function $\sigma(n)^2$. Since by Euler's product $$ \forall s>3,\qquad \sum_{n\geq 1}\frac{\sigma(n)^2}{n^s}=\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)} $$ by tauberian theorems we have $\sum_{n\leq x}\sigma(n)^2 = \frac{5\zeta(3)}{6}x^3+O(x^2)$. On the other hand $3\zeta(4)\approx\zeta(3)\zeta(2)^2$ leads to $\zeta(3)\approx\frac{6}{5}$ and $\sum_{n\leq x}\sigma(n)^2\approx n^3$.