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Fungrim entry: 11302a

π212=n=1(1)n+1n2\frac{{\pi}^{2}}{12} = \sum_{n=1}^{\infty} \frac{{\left(-1\right)}^{n + 1}}{{n}^{2}}
\frac{{\pi}^{2}}{12} = \sum_{n=1}^{\infty} \frac{{\left(-1\right)}^{n + 1}}{{n}^{2}}
Fungrim symbol Notation Short description
Powab{a}^{b} Power
Piπ\pi The constant pi (3.14...)
Sumnf(n)\sum_{n} f(n) Sum
Infinity\infty Positive infinity
Source code for this entry:
    Formula(Equal(Div(Pow(Pi, 2), 12), Sum(Div(Pow(-1, Add(n, 1)), Pow(n, 2)), For(n, 1, Infinity)))))

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2021-03-15 19:12:00.328586 UTC