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Fungrim entry: 6b9f81

1π=229801n=0(4n)!(1103+26390n)(n!)43964n\frac{1}{\pi} = \frac{2 \sqrt{2}}{9801} \sum_{n=0}^{\infty} \frac{\left(4 n\right)! \left(1103 + 26390 n\right)}{{\left(n !\right)}^{4} \cdot {396}^{4 n}}
\frac{1}{\pi} = \frac{2 \sqrt{2}}{9801} \sum_{n=0}^{\infty} \frac{\left(4 n\right)! \left(1103 + 26390 n\right)}{{\left(n !\right)}^{4} \cdot  {396}^{4 n}}
Fungrim symbol Notation Short description
Piπ\pi The constant pi (3.14...)
Sqrtz\sqrt{z} Principal square root
Sumnf(n)\sum_{n} f(n) Sum
Factorialn!n ! Factorial
Powab{a}^{b} Power
Infinity\infty Positive infinity
Source code for this entry:
    Formula(Equal(Div(1, Pi), Mul(Div(Mul(2, Sqrt(2)), 9801), Sum(Div(Mul(Factorial(Mul(4, n)), Add(1103, Mul(26390, n))), Mul(Pow(Factorial(n), 4), Pow(396, Mul(4, n)))), For(n, 0, Infinity))))))

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2020-08-27 09:56:25.682319 UTC