tl:dr at the end. In my first year undergraduate course having definite integrals, I studied about the $\beta$ and $\Gamma$ functions. There was a problem from this particular exam which I never had a short solution to which I revisited after learning about these functions. The problem is roughly as follows (converted to a proof):
$IIT JEE : 2010 : Paper- 1$
Prove:
$I = \int_0^1 x^4*(1-x)^4/(1+x^2)dx = 22/7 - \pi$
Since $|x^2| <1 $ in this interval, I decided to use the binomial expansion for negative indices to reduce this integral to an infinite sum of $\beta$ functions as seen below
$I = \lim_{n->\infty} \int_0^1 x^4(1-x)^4*(1-x^2+x^4-x^6+x^8\ldots(-1)^n *x^{2n})dx$
$I = \lim_{n->\infty} \Sigma_{k=0}^{n}(-1)^k\beta(5 + 2k,5)$
Using $\beta(m,n) = \Gamma(m)* \Gamma(n)/ \Gamma(m+n)$ along with
$\Gamma(m) = (m-1)! :\forall m,n \space\epsilon \space N$
$I = \lim_{n->\infty} \Sigma_{k=0}^{n}(-1)^k(2k+4)!*4!/(2k+9)!$
Now I realised that this isn't going to help me solve this problem but I felt like I came to a discovery I didn't understand. I was wondering why $22/7 - I = \pi$
tl:dr:
Is there an easy way/popular series, or anything neat to show that $22/7 - I = 22/7 - \lim_{n->\infty} \Sigma_{k=0}^{n}(-1)^k(2k+4)!*4!/(2k+9)! = \pi$
or that the exact absolute error between $\pi $ and it's popular approximation $22/7$ is $I = \lim_{n->\infty} \Sigma_{k=0}^{n}(-1)^k(2k+4)!*4!/(2k+9)!$
In order to make my question more clear: I am looking for an easier/elegant way to find this limit $I = \lim_{n->\infty} \Sigma_{k=0}^{n}(-1)^k(2k+4)!*4!/(2k+9)!$