Is there a series to show $22\pi^4>2143\,$?

Thanks to a clever suggestion by J. Lafont, there is a series that can prove $\displaystyle\pi^4>\frac{2143}{22}$. However, it does not use $\pi^8$ but $\pi^{12}$. We start with,

$$\frac{691\pi^{12}}{638512875} = \sum_{k=0}^\infty \frac{1}{(k+1)^{12}}$$

$$\frac{691\pi^{12}}{638668800}-1 = \sum_{k=0}^\infty \frac{1}{(2k+3)^{12}}$$

Multiply them with unknowns $a,b,$ then add the two,

$$\frac{691}{420}\frac{(4096a+4095b)\pi^{12}}{13!}-b = \sum_{k=0}^\infty \left(\frac{a}{(k+1)^{12}}+\frac{b}{(2k+3)^{12}}\right)$$

Let $b=\big(\frac{2143}{22}\big)^3$, and choose $a$ such that $\pi^{12}$ has a unit coefficient. We then get,

$$\begin{aligned}\pi^{12}-\Big(\frac{2143}{22}\Big)^3 &=\sum_{k=0}^\infty\left( -\frac{52410418515}{691\cdot10648\cdot(2k+2)^{12}} +\Big(\frac{2143}{22}\Big)^3\frac{1}{(2k+3)^{12}}\right)\\ &=\sum_{k=0}^\infty R(k) \end{aligned}$$

When expanded out, the coefficients of $R(k)$ are all positive. Thus, the $\text{LHS}$ must be positive. Since it is a difference of two cubes $p^3-q^3 = (p-q)(p^2+pq+q^2)$, then that implies $\displaystyle\pi^4>\frac{2143}{22}.$

From the accelerated series $$\zeta(4)=\frac{\pi^4}{90}=\frac{36}{17}\sum_{n=1}^{\infty } \frac{1}{n^{4}\dbinom{2n}{n}}$$ (Convergence acceleration technique for $\zeta(4)$ (or for $\eta(4)$) via creative telescoping?)

we have the direct sum

$$\pi^4 -\frac{2143}{22}= \frac{5}{52898832} \sum_{n=10}^\infty \left( \frac{1998926767}{n^4\dbinom{2n}{n}}+\frac{17452241}{(n+1)^4\dbinom{2(n+1)}{n+1}}\right)$$

The denominator in the coefficient fraction factors into small primes: $$52898832=(2·3·7)^4·17$$

This series may be related to these.