Pi trochę inaczej niż Leibniz

[Eariu52]

\(\displaystyle{ \frac{ \pi }{2 \sqrt{2} } = \frac{1}{1} + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{13} - \frac{1}{15} + ... }\)

\(\displaystyle{ \frac{ \pi }{2 \sqrt{2} } = \sum_{n=0}^{ \infty } \frac{\left( -1\right) ^{n} }{4n+1} + \frac{\left( -1\right) ^{n} }{4n+3} }\)

Dodano po 7 godzinach 7 minutach 27 sekundach:
Przepraszam, to jest znane, bo jest w OEIS.

Dodano po 8 dniach 16 godzinach 7 minutach 52 sekundach:
Może to nie jest znane:

\(\displaystyle{ \frac{5 \pi }{12} = \frac{1}{1} + \frac{1}{3} + \frac{1}{5} - \frac{1}{7} - \frac{1}{9} - \frac{1}{11} + ... }\)

\(\displaystyle{ \frac{5 \pi }{12} = \sum_{n=0}^{ \infty } \frac{\left( -1\right) ^{n} }{6n+1} + \frac{\left( -1\right) ^{n} }{6n+3} + \frac{\left( -1\right) ^{n} }{6n+5} }\)

Dodano po 5 dniach 23 godzinach 37 minutach 23 sekundach:
\(\displaystyle{ \frac{3 \pi }{8} = \frac{ \sqrt{2} }{1} - \frac{1}{3} + \frac{1}{5} - \frac{ \sqrt{2} }{7} + \frac{ \sqrt{2} }{9} - \frac{1}{11} + \frac{1}{13} - \frac{ \sqrt{2} }{15} + ... }\)

\(\displaystyle{ \frac{3 \pi }{8} = \sum_{n=0}^{ \infty } \frac{ \sqrt{2} }{8n+1} - \frac{1}{8n+3} + \frac{1}{8n+5} - \frac{ \sqrt{2} }{8n+7} }\)

Dodano po 4 miesiącach 20 dniach 37 minutach 47 sekundach:
\(\displaystyle{ \frac{3 \pi }{4} = \frac{2}{1} + \frac{1}{3} + \frac{2}{5} - \frac{2}{7} - \frac{1}{9} - \frac{2}{11} + ... }\)

\(\displaystyle{ \frac{3 \pi }{4} = \sum_{n=0}^{ \infty } \frac{2\left( -1\right) ^{n} }{6n+1} + \frac{\left( -1\right) ^{n} }{6n+3} + \frac{2\left( -1\right) ^{n} }{6n+5} }\)

Dodano po 1 dniu 7 godzinach 19 minutach 53 sekundach:
\(\displaystyle{ \frac{\left( 5k+1\right) \pi }{12} = \sum_{n=0}^{ \infty } \frac{k\left( -1\right) ^{n} }{6n+1} + \frac{\left( k+1\right)\left( -1\right) ^{n} }{6n+3} + \frac{k\left( -1\right) ^{n} }{6n+5} }\)

\(\displaystyle{ \frac{\left( 5k-1\right) \pi }{12} = \sum_{n=0}^{ \infty } \frac{k\left( -1\right) ^{n} }{6n+1} + \frac{\left( k-1\right)\left( -1\right) ^{n} }{6n+3} + \frac{k\left( -1\right) ^{n} }{6n+5} }\)

Dodano po 2 dniach 22 godzinach 17 minutach 13 sekundach:
\(\displaystyle{ \frac{\left( 5k+r\right) \pi }{12} = \sum_{n=0}^{ \infty } \frac{k\left( -1\right) ^{n} }{6n+1} + \frac{\left( k+r\right)\left( -1\right) ^{n} }{6n+3} + \frac{k\left( -1\right) ^{n} }{6n+5} }\)

[Eariu52]

\(\displaystyle{ \frac{5 \pi }{12} = \sum_{n=0}^{ \infty } \frac{ \sqrt{3} }{12n+1} - \frac{ \sqrt{3} }{12n+3} + \frac{1}{12n+5} - \frac{1}{12n+7} + \frac{ \sqrt{3} }{12n+9} - \frac{ \sqrt{3} }{12n+11} }\)

[Eariu52]

\(\displaystyle{ \frac{2 \pi }{3} = \sum_{n=0}^{ \infty } \frac{3 }{12n+1} - \frac{ 2\sqrt{3} }{12n+3} + \frac{1}{12n+5} - \frac{1}{12n+7} + \frac{ 2\sqrt{3} }{12n+9} - \frac{ 3 }{12n+11} }\)

[Eariu52]

\(\displaystyle{ \frac{\left( k+2\right) \pi }{6} = \sum_{n=0}^{ \infty } \frac{k+1}{12n+1} - \frac{k \sqrt{3} }{12n+3} + \frac{1}{12n+5} - \frac{1}{12n+7} + \frac{k \sqrt{3} }{12n+9} - \frac{k+1}{12n+11}, k \ge 1}\)

[Eariu52]

\(\displaystyle{ \frac{\left( 3k+2\right) \pi }{12} = \sum_{n=0}^{ \infty } \frac{k \sqrt{3} }{12n+1} - \frac{\left( 2k-1\right) \sqrt{3}}{12n+3} + \frac{1}{12n+5} - \frac{1}{12n+7} + \frac{\left( 2k-1\right) \sqrt{3}}{12n+9} - \frac{k \sqrt{3} }{12n+11} , k \ge 1 }\)

[Eariu52]

\(\displaystyle{ \frac{\left( 7-k\right) \pi }{12} = \sum_{n=0}^{ \infty } \frac{ \sqrt{3} }{12n+1} - \frac{k}{12n+3} + \frac{2}{12n+5} - \frac{2}{12n+7} + \frac{k}{12n+9} - \frac{ \sqrt{3} }{12n+11}}\)