TITLE: Is there a series to show $22\pi^4>2143\,$?
QUESTION [10 upvotes]: This extends this post.

I. For $\pi^3$:

$$\pi^6-31^2 =\sum_{k=0}^\infty\left(-\frac{63}{(2k+2)^6}+\frac{31^2}{(2k+3)^6}\right) =\sum_{k=0}^\infty P_1(k)\tag1$$
As pointed out by J. Lafont, when $P_1(k)$ is expanded out, its coefficients are all positive. Thus so is the $\text{LHS}$, implying $\pi^3>31$.

II. For $\pi^4$:

The convergents of $\pi^4$ are,
$$97,\, \frac{195}{2},\, \frac{487}{5},\, \frac{1656}{17},\, \frac{2143}{22},\dots$$
The last one, being the particularly close approximation $22\pi^4 \approx 2143.0000027$, was mentioned by Ramanujan. (See also this post.) Using,
$$\frac{\pi^8}{9450}=\sum_{k=0}^\infty \frac{1}{(k+1)^8}$$
$$\frac{17\pi^8}{161280}=\sum_{k=0}^\infty \frac{1}{(2k+1)^8}$$
and the same method to find $(1)$, we get,
$$\pi^8-\Big(\frac{487}{5}\Big)^2 =\sum_{k=0}^\infty\left(\frac{381}{5(2k+2)^8}+\frac{r_1^2}{(2k+3)^8}\right)=\sum_{k=0}^\infty P_2(k)\tag2$$
$$\pi^8-\Big(\frac{2143}{22}\Big)^2 =\sum_{k=0}^\infty\left(-\frac{181695}{11^2(2k+2)^8}+\frac{r_2^2}{(2k+3)^8}\right)=\sum_{k=0}^\infty Q_1(k)\tag3$$
$$\pi^8-\Big(\frac{2143}{22}\Big)^2 =\sum_{k=0}^\infty\left(\frac{r_2^2}{(k+2)^8}-\frac{70208}{1815(2k+1)^8} \right)=\sum_{k=0}^\infty Q_2(k)\tag4$$
where $r_1 =\frac{487}{5},\,$ $r_2 =\frac{2143}{22}$. The coefficients of $P_2(k)$ are all positive, so $5\pi^4>487$. 
However, when the $Q_i(k)$ are expanded out, the constant term for both is negative, so we cannot make an analogous conclusion. (In fact, it takes several terms before the sum turns positive.)

Q: Can one find a similar series for $\pi^8-\Big(\frac{2143}{22}\Big)^2 = \sum_{k=0}^\infty R(k)$ such that all coefficients are positive and immediately implying $22\pi^4>2143$?

REPLY [3 votes]: 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.