他の級数
出典: フリー百科事典『ウィキペディア(Wikipedia)』 (2022/03/26 01:11 UTC 版)
ζ ( 2 ) = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 {\displaystyle \zeta (2)={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}} (バーゼルの問題、リーマンゼータ関数) ζ ( 4 ) = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ = π 4 90 {\displaystyle \zeta (4)={\frac {1}{1^{4}}}+{\frac {1}{2^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{4^{4}}}+\cdots ={\frac {\pi ^{4}}{90}}} ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = 1 1 2 n + 1 2 2 n + 1 3 2 n + 1 4 2 n + ⋯ = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}\,={\frac {1}{1^{2n}}}+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots =(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} , ただし B2n は ベルヌーイ数) ∑ n = 1 ∞ 3 n − 1 4 n ζ ( n + 1 ) = π {\displaystyle \sum _{n=1}^{\infty }{\frac {3^{n}-1}{4^{n}}}\,\zeta (n+1)=\pi } ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 1 = 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = arctan 1 = π 4 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{1}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots =\arctan {1}={\frac {\pi }{4}}} (ライプニッツ公式) ∑ n = 1 ∞ ( − 1 ) n + 1 n 2 = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯ = π 2 12 {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{2}}}={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{12}}} ∑ n = 1 ∞ 1 ( 2 n ) 2 = 1 2 2 + 1 4 2 + 1 6 2 + 1 8 2 + ⋯ = π 2 24 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n)^{2}}}={\frac {1}{2^{2}}}+{\frac {1}{4^{2}}}+{\frac {1}{6^{2}}}+{\frac {1}{8^{2}}}+\cdots ={\frac {\pi ^{2}}{24}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 2 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = π 2 8 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{2}={\frac {1}{1^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}+\cdots ={\frac {\pi ^{2}}{8}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 3 = 1 1 3 − 1 3 3 + 1 5 3 − 1 7 3 + ⋯ = π 3 32 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{3}={\frac {1}{1^{3}}}-{\frac {1}{3^{3}}}+{\frac {1}{5^{3}}}-{\frac {1}{7^{3}}}+\cdots ={\frac {\pi ^{3}}{32}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 4 = 1 1 4 + 1 3 4 + 1 5 4 + 1 7 4 + ⋯ = π 4 96 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{4}={\frac {1}{1^{4}}}+{\frac {1}{3^{4}}}+{\frac {1}{5^{4}}}+{\frac {1}{7^{4}}}+\cdots ={\frac {\pi ^{4}}{96}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 5 = 1 1 5 − 1 3 5 + 1 5 5 − 1 7 5 + ⋯ = 5 π 5 1536 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{5}={\frac {1}{1^{5}}}-{\frac {1}{3^{5}}}+{\frac {1}{5^{5}}}-{\frac {1}{7^{5}}}+\cdots ={\frac {5\pi ^{5}}{1536}}} ∑ n = 0 ∞ ( ( − 1 ) n 2 n + 1 ) 6 = 1 1 6 + 1 3 6 + 1 5 6 + 1 7 6 + ⋯ = π 6 960 {\displaystyle \sum _{n=0}^{\infty }{\left({\frac {(-1)^{n}}{2n+1}}\right)}^{6}={\frac {1}{1^{6}}}+{\frac {1}{3^{6}}}+{\frac {1}{5^{6}}}+{\frac {1}{7^{6}}}+\cdots ={\frac {\pi ^{6}}{960}}} ∑ n = 0 ∞ 1 ( 4 n + 1 ) ( 4 n + 3 ) = 1 1 ⋅ 3 + 1 5 ⋅ 7 + 1 9 ⋅ 11 + ⋯ = π 8 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(4n+1)(4n+3)}}={\frac {1}{1\cdot 3}}+{\frac {1}{5\cdot 7}}+{\frac {1}{9\cdot 11}}+\cdots ={\frac {\pi }{8}}} π = 1 + 1 2 + 1 3 + 1 4 − 1 5 + 1 6 + 1 7 + 1 8 + 1 9 − 1 10 + 1 11 + 1 12 − 1 13 + ⋯ {\displaystyle \pi ={1}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}-{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}-{\frac {1}{10}}+{\frac {1}{11}}+{\frac {1}{12}}-{\frac {1}{13}}+\cdots } (オイラー、1748年) この式では、最初の2つの項の後、符号は次のように決定される。分母が4m - 1で表される素数である場合、符号は正であり。分母が4m + 1で表される素数である場合、符号は負である。 合成数の場合、符号はその素因数分解した素数の符号の積に等しい。 また ∑ n = 1 ∞ F 2 n n 2 ( 2 n n ) = 4 π 2 25 5 {\displaystyle \sum _{n=1}^{\infty }{\frac {F_{2n}}{n^{2}{\binom {2n}{n}}}}={\frac {4\pi ^{2}}{25{\sqrt {5}}}}} ただし F n {\displaystyle F_{n}} はn番目のフィボナッチ数。
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