[math-fun] Happy approximate pi day
I see from various friends that it is approximate pi day, 22/7 if you live outside the USA. I’m sure many of us know Dalzell’s classic result that int_0^1 x^4(1-x)^4/(1+x^2) dx = 22/7 - pi which proves that pi is strictly less than 22/7, since the integrand is nonnegative. But for fun, here are a couple of infinite series with positive terms that also equal 22/7 - pi: sum_{n=2}^\infty 60/[(4*n^2-1)(16*n^2-1)(16*n^2-9)] sum_{n=1}^\infty 640n/[(4n+1)(4n+3)(4n+5)(4n+7)(4n+9)(4n+11)] sum_{n=6}^infty 2^{n-3}(n+242)/[15(2n+1)(2n+3)C(2n,n)] Stege
Also, Archimedes proved 22/7 > Pi. Beukers discusses the 22/7 integral as one of many similar approximations, see: https://oeis.org/A123178 https://oeis.org/A006139 and references therein. Regarding Beukers/Hata integral theory, is there a no-go theorem saying that, infinitely many, sufficient quality rational approximants *can not* be produced by integrating entire holomorphic functions only? --Brad On Mon, Jul 22, 2019 at 2:48 PM Lucas, Stephen K - lucassk <lucassk@jmu.edu> wrote:
I see from various friends that it is approximate pi day, 22/7 if you live outside the USA. I’m sure many of us know Dalzell’s classic result that
int_0^1 x^4(1-x)^4/(1+x^2) dx = 22/7 - pi
which proves that pi is strictly less than 22/7, since the integrand is nonnegative. But for fun, here are a couple of infinite series with positive terms that also equal 22/7 - pi:
sum_{n=2}^\infty 60/[(4*n^2-1)(16*n^2-1)(16*n^2-9)]
sum_{n=1}^\infty 640n/[(4n+1)(4n+3)(4n+5)(4n+7)(4n+9)(4n+11)]
sum_{n=6}^infty 2^{n-3}(n+242)/[15(2n+1)(2n+3)C(2n,n)]
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participants (2)
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Brad Klee -
Lucas, Stephen K - lucassk