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1 l o g ( n ) = ∑ k = 1 ∞ ( 1 k − 1 n ⌈ k n ⌉ ) {\displaystyle log(n)=\sum _{k=1}^{\infty }\left({\frac {1}{k}}-{\frac {1}{n\lceil {\frac {k}{n}}\rceil }}\right)}
2 l o g ( p ) = − ∑ n = 1 ∞ 1 n ∑ k = 1 p − 1 e 2 π i k n p {\displaystyle log(p)=-\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{k=1}^{p-1}e^{2\pi i{\frac {kn}{p}}}}
A167407
3 l o g ( p ) = lim n → ∞ ∑ k = n + 1 n p 1 k {\displaystyle log(p)=\lim _{n\to \infty }\sum _{k=n+1}^{np}{\frac {1}{k}}}
(Eric Naslund answered
l o g ( n ) = . . . = lim M → ∞ ∑ k = 1 n M 1 k − ∑ k = 1 M 1 k {\displaystyle log(n)=...=\lim _{M\to \infty }\sum _{k=1}^{nM}{\frac {1}{k}}-\sum _{k=1}^{M}{\frac {1}{k}}}
to this question by Mats Granvik)