TITLE: A magnificent series for $\pi-333/106$
QUESTION [24 upvotes]: Stated here without proof is the magnificent series $$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106},$$ which proves that $\pi>333/106$. 

I can only assume that the series is proven using the integral $$\pi-\frac{333}{106}=\frac{1}{530}\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{1+x^2}dx.$$My attempts have been so far to split up the integral as
$$\begin{align}
530J&=\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{1+x^2}dx\\
&=197\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx+462\int_0^1\frac{x^{7}(1-x)^6}{1+x^2}dx\\
&=197J_1+462J_2.
\end{align}$$
Each remaining integral is turned into a series with 
$$\frac1{1+x^2}=\sum_{n\ge0}(-1)^n x^{2n}$$
so we have two series of the form
$$f(p)=\sum_{n\ge0}(-1)^n\int_0^1 x^{p+2n}(1-x)^6dx=720\sum_{n\ge0}(-1)^n\frac{(p+2n)!}{(p+2n+7)!}.$$
Each factorial term is rewritten as $$\frac{s!}{(s+7)!}=\frac1{(s+1)(s+2)(s+3)(s+4)(s+5)(s+6)(s+7)},$$
so that 
$$f(p)=720\sum_{n\ge0}\frac{(-1)^n}{\prod_{k=1}^{7}(2n+p+k)}.$$
Then $$J_1=f(5)\\ J_2=f(7).$$
But how does one get from $530J=197f(5)+462f(7)$ to the series in question? Furthermore, how do we prove that $J=\pi-333/106$? I would assume that one would at some point use the binomial theorem then be left with a bunch of integrals like $$\int_0^1\frac{x^qdx}{1+x^2}$$ which I suppose are evaluable in terms of $\pi$, but it seems like a great deal of cancellation/simplification would have to occur and I do not immediately see where this would happen. There has to be an easier way. 
Thanks!

REPLY [8 votes]: This series can be obtained with the same technique used in A series to prove $\frac{22}{7}-\pi>0$
Let us start from series
$$\sum _{k=0}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{64}{21}$$ 
to obtain the following truncations:
$$\sum _{k=1}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{2176}{693}$$
$$\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{4288}{1365}$$
$$\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\pi-\frac{45708032}{14549535}$$ 
The approximation we are interested in lies between two of these fractions. 
$$ \frac{4288}{1365}< \frac{333}{106} < \frac{45708032}{14549535}$$ 
Therefore, a series for $\pi-\frac{333}{106}$ can be obtained as a mix of the series for $\pi-\frac{4288}{1365}$ and $\pi-\frac{45708032}{14549535}$. 
From $$\pi-\frac{333}{106} = a(\pi-\frac{4288}{1365})+b(\pi-\frac{45708032}{14549535})$$
we obtain 
$$a=\frac{56237}{237440}$$
$$b=\frac{181203}{237440}$$
Finally,
$$\pi-\frac{333}{106}=\frac{56237}{237440}(\pi-\frac{4288}{1365})+\frac{181203}{237440}(\pi-\frac{45708032}{14549535})=$$
$$\frac{56237}{237440}\sum _{k=2}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}+$$
$$\frac{181203}{237440}\sum _{k=3}^\infty \frac{960}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=$$
$$\frac{56237}{237440}\sum _{k=0}^\infty \frac{960}{(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)(4k+19)}+$$
$$\frac{181203}{237440}\sum _{k=0}^\infty \frac{960}{(4k+13)(4k+15)(4k+17)(4k+19)(4k+21)(4k+23)}=$$
$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)}$$