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歐拉-馬斯刻若尼常數 歐拉-馬斯刻若尼常數 藍色區域的面積收斂到歐拉常數
符號 γ {\displaystyle \gamma } 位數 數列編號 A001620 定義 γ = lim n → ∞ [ ( ∑ k = 1 n 1 k ) − ln ( n ) ] {\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]} γ = ∫ 1 ∞ ( 1 ⌊ x ⌋ − 1 x ) d x {\displaystyle \gamma =\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx} 連分數 [0; 1, 1, 2, 1, 2, 1, 4, 3, 13, 5, 1, 1, 8, 1, 2, 4, 1, 1, 40, ...] 值 γ ≈ {\displaystyle \gamma \approx } 無窮級數 γ = ∑ k = 1 ∞ [ 1 k − ln ( 1 + 1 k ) ] {\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]} 二进制 0.10010011 1100 0100 0110 0111 … 十进制 0.57721566 4901 5328 6060 6512 … 十六进制 0.93C467E3 7DB0 C7A4 D1BE 3F81 …
歐拉-馬斯刻若尼常數 是一个数学常数 ，定义为调和级数 与自然对数 的差值：
γ = lim n → ∞ [ ( ∑ k = 1 n 1 k ) − ln ( n ) ] = ∫ 1 ∞ ( 1 ⌊ x ⌋ − 1 x ) d x {\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx} 它的近似值为 γ ≈ 0.577215664901532860606512090082402431042159335 {\displaystyle \gamma \approx 0.577215664901532860606512090082402431042159335} [ 1]
歐拉-馬斯刻若尼常數主要应用于数论 。
该常数最先由瑞士 数学家莱昂哈德·欧拉 在1735年发表的文章De Progressionibus harmonicus observationes 中定义。欧拉曾经使用 C {\displaystyle C} 意大利 数学家洛倫佐·馬斯凱羅尼 引入了 γ {\displaystyle \gamma }
目前尚不知道该常数是否为有理数 ，但是分析表明如果它是一个有理数，那么它的分母位数将超过10242080 。[ 2]
− γ = Γ ′ ( 1 ) = Ψ ( 1 ) {\displaystyle \ -\gamma =\Gamma '(1)=\Psi (1)} γ = lim x → ∞ [ x − Γ ( 1 x ) ] {\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]} γ = lim n → ∞ [ Γ ( 1 n ) Γ ( n + 1 ) n 1 + 1 n Γ ( 2 + n + 1 n ) − n 2 n + 1 ] {\displaystyle \gamma =\lim _{n\to \infty }\left[{\frac {\Gamma ({\frac {1}{n}})\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma (2+n+{\frac {1}{n}})}}-{\frac {n^{2}}{n+1}}\right]} γ = ∑ m = 2 ∞ ( − 1 ) m ζ ( m ) m {\displaystyle \gamma =\sum _{m=2}^{\infty }{\frac {(-1)^{m}\zeta (m)}{m}}} = ln ( 4 π ) + ∑ m = 1 ∞ ( − 1 ) m − 1 ζ ( m + 1 ) 2 m ( m + 1 ) {\displaystyle =\ln \left({\frac {4}{\pi }}\right)+\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}\zeta (m+1)}{2^{m}(m+1)}}} lim ε → 0 ζ ( 1 + ε ) + ζ ( 1 − ε ) 2 = γ {\displaystyle \lim _{\varepsilon \to 0}{\frac {\zeta (1+\varepsilon )+\zeta (1-\varepsilon )}{2}}=\gamma } γ = 3 2 − ln 2 − ∑ m = 2 ∞ ( − 1 ) m m − 1 m [ ζ ( m ) − 1 ] {\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]} = lim n → ∞ [ 2 n − 1 2 n − ln n + ∑ k = 2 n ( 1 k − ζ ( 1 − k ) n k ) ] {\displaystyle =\lim _{n\to \infty }\left[{\frac {2\,n-1}{2\,n}}-\ln \,n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right]} = lim n → ∞ [ 2 n e 2 n ∑ m = 0 ∞ 2 m n ( m + 1 ) ! ∑ t = 0 m 1 t + 1 − n ln 2 + O ( 1 2 n e 2 n ) ] {\displaystyle =\lim _{n\to \infty }\left[{\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{m\,n}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\,\ln 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right]} γ = lim s → 1 + ∑ n = 1 ∞ ( 1 n s − 1 s n ) = lim s → 1 + ( ζ ( s ) − 1 s − 1 ) {\displaystyle \gamma =\lim _{s\to 1^{+}}\sum _{n=1}^{\infty }\left({\frac {1}{n^{s}}}-{\frac {1}{s^{n}}}\right)=\lim _{s\to 1^{+}}\left(\zeta (s)-{\frac {1}{s-1}}\right)} γ = lim x → ∞ [ x − Γ ( 1 x ) ] {\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]} = lim n → ∞ 1 n ∑ k = 1 n ( ⌈ n k ⌉ − n k ) {\displaystyle =\lim _{n\to \infty }{\frac {1}{n}}\,\sum _{k=1}^{n}\left(\left\lceil {\frac {n}{k}}\right\rceil -{\frac {n}{k}}\right)} γ = ∑ k = 1 n 1 k − ln ( n ) − ∑ m = 2 ∞ ζ ( m , n + 1 ) m {\displaystyle \gamma =\sum _{k=1}^{n}{\frac {1}{k}}-\ln(n)-\sum _{m=2}^{\infty }{\frac {\zeta (m,n+1)}{m}}} γ = − ∫ 0 ∞ e − x ln x d x = ∫ ∞ 0 e − x ln x d x {\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx=\int _{\infty }^{0}{e^{-x}\ln x}\,dx} [ 證明 1] = − ∫ 0 1 ln ln 1 x d x {\displaystyle =-\int _{0}^{1}{\ln \ln {\frac {1}{x}}}\,dx} = ∫ 0 ∞ ( 1 1 − e − x − 1 x ) e − x d x {\displaystyle =\int _{0}^{\infty }{\left({\frac {1}{1-e^{-x}}}-{\frac {1}{x}}\right)e^{-x}}\,dx} = ∫ 0 ∞ 1 x ( 1 1 + x − e − x ) d x {\displaystyle =\int _{0}^{\infty }{{\frac {1}{x}}\left({\frac {1}{1+x}}-e^{-x}\right)}\,dx} ∫ 0 ∞ e − x 2 ln x d x = − 1 4 ( γ + 2 ln 2 ) π {\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\ln x}\,dx=-{\tfrac {1}{4}}(\gamma +2\ln 2){\sqrt {\pi }}} ∫ 0 ∞ e − x ln 2 x d x = γ 2 + π 2 6 {\displaystyle \int _{0}^{\infty }{e^{-x}\ln ^{2}x}\,dx=\gamma ^{2}+{\frac {\pi ^{2}}{6}}} γ = ∫ 0 1 ∫ 0 1 x − 1 ( 1 − x y ) ln ( x y ) d x d y = ∑ n = 1 ∞ ( 1 n − ln n + 1 n ) {\displaystyle \gamma =\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-x\,y)\ln(x\,y)}}\,dx\,dy=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)} ∑ n = 1 ∞ N 1 ( n ) + N 0 ( n ) 2 n ( 2 n + 1 ) = γ {\displaystyle \sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}=\gamma } γ = ∑ k = 1 ∞ [ 1 k − ln ( 1 + 1 k ) ] {\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]} γ = 1 − ∑ k = 2 ∞ ( − 1 ) k ⌊ log 2 k ⌋ k + 1 {\displaystyle \gamma =1-\sum _{k=2}^{\infty }(-1)^{k}{\frac {\lfloor \log _{2}k\rfloor }{k+1}}}
γ = ∑ k = 2 ∞ ( − 1 ) k ⌊ log 2 k ⌋ k = 1 2 − 1 3 + 2 ( 1 4 − 1 5 + 1 6 − 1 7 ) + 3 ( 1 8 − ⋯ − 1 15 ) + … {\displaystyle \gamma =\sum _{k=2}^{\infty }(-1)^{k}{\frac {\left\lfloor \log _{2}k\right\rfloor }{k}}={\tfrac {1}{2}}-{\tfrac {1}{3}}+2\left({\tfrac {1}{4}}-{\tfrac {1}{5}}+{\tfrac {1}{6}}-{\tfrac {1}{7}}\right)+3\left({\tfrac {1}{8}}-\dots -{\tfrac {1}{15}}\right)+\dots } γ + ζ ( 2 ) = ∑ k = 1 ∞ 1 k ⌊ k ⌋ 2 = 1 + 1 2 + 1 3 + 1 4 ( 1 4 + ⋯ + 1 8 ) + 1 9 ( 1 9 + ⋯ + 1 15 ) + … {\displaystyle \gamma +\zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k\lfloor {\sqrt {k}}\rfloor ^{2}}}=1+{\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{4}}\left({\tfrac {1}{4}}+\dots +{\tfrac {1}{8}}\right)+{\tfrac {1}{9}}\left({\tfrac {1}{9}}+\dots +{\tfrac {1}{15}}\right)+\dots }
γ = ∑ k = 2 ∞ k − ⌊ k ⌋ 2 k 2 ⌊ k ⌋ 2 = 1 2 2 + 2 3 2 + 1 2 2 ( 1 5 2 + 2 6 2 + 3 7 2 + 4 8 2 ) + 1 3 2 ( 1 10 2 + ⋯ + 6 15 2 ) + … {\displaystyle \gamma =\sum _{k=2}^{\infty }{\frac {k-\lfloor {\sqrt {k}}\rfloor ^{2}}{k^{2}\lfloor {\sqrt {k}}\rfloor ^{2}}}={\tfrac {1}{2^{2}}}+{\tfrac {2}{3^{2}}}+{\tfrac {1}{2^{2}}}\left({\tfrac {1}{5^{2}}}+{\tfrac {2}{6^{2}}}+{\tfrac {3}{7^{2}}}+{\tfrac {4}{8^{2}}}\right)+{\tfrac {1}{3^{2}}}\left({\tfrac {1}{10^{2}}}+\dots +{\tfrac {6}{15^{2}}}\right)+\dots }
γ = ∫ 0 1 1 1 + x ∑ n = 1 ∞ x 2 n − 1 d x {\displaystyle \gamma =\int _{0}^{1}{\frac {1}{1+x}}\sum _{n=1}^{\infty }x^{2^{n}-1}\,dx} γ {\displaystyle \gamma } 连分数 展开式为：
γ = [ 0 ; 1 , 1 , 2 , 1 , 2 , 1 , 4 , 3 , 13 , 5 , 1 , 1 , 8 , 1 , 2 , 4 , 1 , 1 , 40 , . . . ] {\displaystyle \gamma =[0;1,1,2,1,2,1,4,3,13,5,1,1,8,1,2,4,1,1,40,...]\,} OEIS 數列A002852 ）. γ ≈ H n − ln ( n ) − 1 2 n + 1 12 n 2 − 1 120 n 4 + . . . {\displaystyle \gamma \approx H_{n}-\ln \left(n\right)-{\frac {1}{2n}}+{\frac {1}{12n^{2}}}-{\frac {1}{120n^{4}}}+...} γ ≈ H n − ln ( n + 1 2 + 1 24 n − 1 48 n 3 + . . . ) {\displaystyle \gamma \approx H_{n}-\ln \left({n+{\frac {1}{2}}+{\frac {1}{24n}}-{\frac {1}{48n^{3}}}+...}\right)} γ ≈ H n − ln ( n ) + ln ( n + 1 ) 2 − 1 6 n ( n + 1 ) + 1 30 n 2 ( n + 1 ) 2 − . . . {\displaystyle \gamma \approx H_{n}-{\frac {\ln \left(n\right)+\ln \left({n+1}\right)}{2}}-{\frac {1}{6n\left({n+1}\right)}}+{\frac {1}{30n^{2}\left({n+1}\right)^{2}}}-...} γ {\displaystyle {\boldsymbol {\gamma }}} 的已知位数 日期 位数 计算者 1734年 5 莱昂哈德·欧拉 1736年 15 莱昂哈德·欧拉 1790年 19 洛倫佐·馬斯凱羅尼 1809年 24 Johann G. von Soldner 1812年 40 F.B.G. Nicolai 1861年 41 Oettinger 1869年 59 William Shanks 1871年 110 William Shanks 1878年 263 约翰·柯西·亚当斯 1962年 1,271 高德纳 1962年 3,566 D.W. Sweeney 1977年 20,700 Richard P. Brent 1980年 30,100 Richard P. Brent 和埃德温·麦克米伦 1993年 172,000 Jonathan Borwein 1997年 1,000,000 Thomas Papanikolaou 1998年12月 7,286,255 Xavier Gourdon 1999年10月 108,000,000 Xavier Gourdon和Patrick Demichel 2006年7月16日 2,000,000,000 Shigeru Kondo和Steve Pagliarulo 2006年12月8日 116,580,041 Alexander J. Yee 2007年7月15日 5,000,000,000 Shigeru Kondo和Steve Pagliarulo 2008年1月1日 1,001,262,777 Richard B. Kreckel 2008年1月3日 131,151,000 Nicholas D. Farrer 2008年6月30日 10,000,000,000 Shigeru Kondo和Steve Pagliarulo 2009年1月18日 14,922,244,771 Alexander J. Yee和Raymond Chan 2009年3月13日 29,844,489,545 Alexander J. Yee和Raymond Chan 2013年 119,377,958,182 Alexander J. Yee 2016年 160,000,000,000 Peter Trueb 2016年 250,000,000,000 Ron Watkins 2017年 477,511,832,674 Ron Watkins 2020年 600,000,000,100 Seungmin Kim和Ian Cutress
^ γ = − ∫ 0 ∞ e − x ln x d x {\displaystyle \gamma =-\int _{0}^{\infty }{e^{-x}\ln x}\,dx} ∫ k k + 1 1 x d x < 1 k < ∫ k − 1 k 1 x d x {\displaystyle \int _{k}^{k+1}{\frac {1}{x}}\,dx<{\frac {1}{k}}<\int _{k-1}^{k}{\frac {1}{x}}\,dx} ∫ k k − 1 1 x d x − 1 k < 1 k ( k − 1 ) {\displaystyle \int _{k}^{k-1}{\frac {1}{x}}\,dx-{\frac {1}{k}}<{\frac {1}{k(k-1)}}} ln n < ∑ k = 1 n 1 k < 1 + ln n {\displaystyle \ln n<\sum _{k=1}^{n}{\frac {1}{k}}<1+\ln n} γ {\displaystyle \gamma } ∑ k = 1 n 1 k {\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}} = ∑ k = 1 n ∫ 0 1 t k − 1 d t {\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}t^{k-1}\,dt} = ∫ 0 1 ∑ k = 1 n t k − 1 d t {\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}t^{k-1}\,dt} = ∫ 0 1 1 − t n 1 − t d t {\displaystyle =\int _{0}^{1}{\frac {1-t^{n}}{1-t}}\,dt} = ∫ n 0 1 − ( 1 − x n ) n 1 − ( 1 − x n ) d ( 1 − x n ) {\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{1-\left(1-{\frac {x}{n}}\right)}}d\left(1-{\tfrac {x}{n}}\right)} = ∫ n 0 1 − ( 1 − x n ) n x n ( − 1 n ) d x {\displaystyle =\int _{n}^{0}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{\frac {x}{n}}}\left(-{\frac {1}{n}}\right)dx} = ∫ 0 n 1 − ( 1 − x n ) n x d x {\displaystyle =\int _{0}^{n}{\frac {1-\left(1-{\frac {x}{n}}\right)^{n}}{x}}dx} ln n = ∫ 1 n 1 x d x , {\displaystyle \ln n=\int _{1}^{n}{\frac {1}{x}}\,dx,} γ = lim n → ∞ ( ∑ k = 1 n 1 k − ln n ) {\displaystyle \gamma =\lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)} = lim n → ∞ [ ∫ 0 n 1 − ( 1 − x / n ) n x d x − ∫ 1 n 1 x d x ] {\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]} = lim n → ∞ [ ∫ 0 1 1 − ( 1 − x / n ) n x d x − ∫ 1 n ( 1 − x / n ) n x ] {\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]} = ∫ 0 1 1 − lim n → ∞ ( 1 − x / n ) n x d x − ∫ 1 ∞ lim n → ∞ ( 1 − x / n ) n x {\displaystyle =\int _{0}^{1}{\frac {1-\lim _{n\to \infty }(1-x/n)^{n}}{x}}\,dx-\int _{1}^{\infty }{\frac {\lim _{n\to \infty }(1-x/n)^{n}}{x}}} = ∫ 0 1 1 − e − x x d x − ∫ 1 ∞ e − x x {\displaystyle =\int _{0}^{1}{\frac {1-e^{-x}}{x}}\,dx-\int _{1}^{\infty }{\frac {e^{-x}}{x}}} = ( 1 − e − x ) ln x | 0 1 − ∫ 0 1 ln x d ( 1 − e − x ) − e − x ln x | 1 ∞ + ∫ 1 ∞ ln x d e − x {\displaystyle =\left.(1-e^{-x})\ln x\right|_{0}^{1}-\int _{0}^{1}\ln x\,d(1-e^{-x})-\left.e^{-x}\ln x\right|_{1}^{\infty }+\int _{1}^{\infty }\ln x\,de^{-x}} = − ∫ 0 ∞ e − x ln x d x . {\displaystyle =-\int _{0}^{\infty }e^{-x}\ln x\,dx.} 前面的放缩法主要是证明了
[ ( ∑ k = 1 n 1 k ) − ln ( n ) ] {\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]} ^ A001620 oeis.org [2014-7-17]^ Havil 2003 p 97. Borwein, Jonathan M., David M. Bradley, Richard E. Crandall. Computational Strategies for the Riemann Zeta Function (PDF) . Journal of Computational and Applied Mathematics. 2000, 121 : 11 [2014-07-17 ] . doi:10.1016/s0377-0427(00)00336-8 原始内容 (PDF) 存档于2006-09-25）. Gourdon, Xavier, and Sebah, P. (2002) "Collection of formulas for Euler's constant, γ. （页面存档备份 ，存于互联网档案馆 ）" Gourdon, Xavier, and Sebah, P. (2004) "The Euler constant: γ. （页面存档备份 ，存于互联网档案馆 ）" Donald Knuth (1997) The Art of Computer Programming , Vol. 1ISBN 978-0-201-89683-1 Krämer, Stefan (2005) Die Eulersche Konstante γ und verwandte Zahlen . Diplomarbeit, Universität Göttingen. Sondow, Jonathan (1998) "An antisymmetric formula for Euler's constant, " Mathematics Magazine 71Sondow, Jonathan (2002) "A hypergeometric approach, via linear forms involving logarithms, to irrationality criteria for Euler's constant. " With an Appendix by Sergey Zlobin , Mathematica Slovaca 59: 307-314. Sondow, Jonathan. An infinite product for eγ via hypergeometric formulas for Euler's constant, γ. 2003. arXiv:math.CA/0306008 Sondow, Jonathan (2003a) "Criteria for irrationality of Euler's constant, " Proceedings of the American Mathematical Society 131Sondow, Jonathan (2005) "Double integrals for Euler's constant and ln 4/π and an analog of Hadjicostas's formula, " American Mathematical Monthly 112Sondow, Jonathan (2005) "New Vacca-type rational series for Euler's constant and its 'alternating' analog ln 4/π. "Sondow, Jonathan; Zudilin, Wadim. Euler's constant, q-logarithms, and formulas of Ramanujan and Gosper. 2006. arXiv:math.NT/0304021 G. Vacca (1926), "Nuova serie per la costante di Eulero, C = 0,577…". Rendiconti, Accademia Nazionale dei Lincei, Roma, Classe di Scienze Fisiche, Matematiche e Naturali (6) 3, 19–20. James Whitbread Lee Glaisher (1872), "On the history of Euler's constant". Messenger of Mathematics. New Series, vol.1, p. 25-30, JFM 03.0130.01Carl Anton Bretschneider (1837). "Theoriae logarithmi integralis lineamenta nova". Crelle Journal, vol.17, p. 257-285 (submitted 1835) Lorenzo Mascheroni (1790). "Adnotationes ad calculum integralem Euleri, in quibus nonnulla problemata ab Eulero proposita resolvuntur". Galeati, Ticini.Lorenzo Mascheroni (1792). "Adnotationes ad calculum integralem Euleri. In quibus nonnullae formulae ab Eulero propositae evolvuntur". Galeati, Ticini. Both online at: http://books.google.de/books?id=XkgDAAAAQAAJ （页面存档备份 ，存于互联网档案馆 ）Havil, Julian. Gamma: Exploring Euler's Constant . Princeton University Press. 2003. ISBN 0-691-09983-9 Karatsuba, E. A. Fast evaluation of transcendental functions. Probl. Inf. Transm. 1991, 27 (44): 339–360. E.A. Karatsuba, On the computation of the Euler constant γ, J. of Numerical Algorithms Vol.24, No.1-2, pp. 83–97 (2000) M. Lerch, Expressions nouvelles de la constante d'Euler. Sitzungsberichte der Königlich Böhmischen Gesellschaft der Wissenschaften 42, 5 p. (1897) Lagarias, Jeffrey C. Euler's constant: Euler's work and modern developments. arXiv:1303.1856 Bulletin of the American Mathematical Society 50 (4): 527-628 (2013)
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提示：此条目的主题不是尤拉數。
A001620
![{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7af12010d95e43a1b424a6c3a76c92c2727c1c06)
![{\displaystyle \gamma =\sum _{k=1}^{\infty }\left[{\frac {1}{k}}-\ln \left(1+{\frac {1}{k}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7176d3dc104a5ef96f1039306f330096dde13ccd)
![{\displaystyle \gamma =\lim _{n\rightarrow \infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]=\int _{1}^{\infty }\left({1 \over \lfloor x\rfloor }-{1 \over x}\right)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cf185b6d9ad97389a24d868420ea2379a75f60a)
![{\displaystyle \gamma =\lim _{x\to \infty }\left[x-\Gamma \left({\frac {1}{x}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6598511fab1a65e9b06953c415d96ee8633c8f2)
![{\displaystyle \gamma =\lim _{n\to \infty }\left[{\frac {\Gamma ({\frac {1}{n}})\Gamma (n+1)\,n^{1+{\frac {1}{n}}}}{\Gamma (2+n+{\frac {1}{n}})}}-{\frac {n^{2}}{n+1}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fc69773463c0fce9d5e859a34842d25a0da6487)
![{\displaystyle \gamma ={\frac {3}{2}}-\ln 2-\sum _{m=2}^{\infty }(-1)^{m}\,{\frac {m-1}{m}}[\zeta (m)-1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87c6c8e2b81d1f232a24da48013a831eb1e9192d)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2\,n-1}{2\,n}}-\ln \,n+\sum _{k=2}^{n}\left({\frac {1}{k}}-{\frac {\zeta (1-k)}{n^{k}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2dc45371ac6aceac57f05f5412ed5e33c4a69d2)
![{\displaystyle =\lim _{n\to \infty }\left[{\frac {2^{n}}{e^{2^{n}}}}\sum _{m=0}^{\infty }{\frac {2^{m\,n}}{(m+1)!}}\sum _{t=0}^{m}{\frac {1}{t+1}}-n\,\ln 2+O\left({\frac {1}{2^{n}\,e^{2^{n}}}}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68a0383cf4feb0933d76bad33d30d5077a663d96)
![{\displaystyle \gamma =[0;1,1,2,1,2,1,4,3,13,5,1,1,8,1,2,4,1,1,40,...]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bd27ebd870468dbb92cf355e1d9525467520096)
![{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{n}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {1}{x}}\,dx\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ad51b5b4b416e2d6f007f02e8c37f90b3f3ea93)
![{\displaystyle =\lim _{n\to \infty }\left[\int _{0}^{1}{\frac {1-(1-x/n)^{n}}{x}}\,dx-\int _{1}^{n}{\frac {(1-x/n)^{n}}{x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f0df6992c0f206f2e4328fca6798cb2fa271768)
![{\displaystyle \left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\ln(n)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54edfd70bc01e72b44b1a850ab7efad0852c82ee)
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