■完全ベキ乗数列(その38)

  1−1/2+1/3−1/4+・・・=log2

  1−1/3+1/5−1/7+・・・=π/4

 それでは,

[Q]1−1/4+1/7−1/10+・・・=?

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[A]1−1/4+1/7−1/10+・・・=π/3√3+1/3・log2

  1−1/2+1/3−1/4+・・・=log2

 1/2・Σ{1/(n+1/2)−1/(n+1)}

  1−1/3+1/5−1/7+・・・=π/4

  1/4・Σ{1/(n+1/4)−1/(n+1)}

 −1/4・Σ{1/(n+3/4)−1/(n+1)}

  1−1/4+1/7−1/10+・・・=?

  1/6・Σ{1/(n+1/6)−1/(n+1)}

 −1/6・Σ{1/(n+4/6)−1/(n+1)}

と書くことができる.

 一般に

  Σ{1/(n+p/q)−1/(n+1)}

=π/2・cotpπ/q+log2q−2Σcos2pkπ/q・logsinkπ/q  (0<k<q/2)

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[1]p=1,q=2

  Σ{1/(n+p/q)−1/(n+1)}=log4

  1/2・Σ{1/(n+1/2)−1/(n+1)}=log2

[2]p=1,q=4

  Σ{1/(n+p/q)−1/(n+1)}=π/2+log8

[3]p=3,q=4

  Σ{1/(n+p/q)−1/(n+1)}=−π/2+log8

  1/4・Σ{1/(n+1/4)−1/(n+1)}

 −1/4・Σ{1/(n+3/4)−1/(n+1)}=π/4

[4]p=1,q=6

  Σ{1/(n+p/q)−1/(n+1)}=π/2・√3+log12−2{1/2・log1/2−1/2・log√3/2}

[5]p=4,q=6

  Σ{1/(n+p/q)−1/(n+1)}=−π/2√3+log12−2{−1/2・log1/2−1/2・log√3/2}

  1/6・Σ{1/(n+1/6)−1/(n+1)}

 −1/6・Σ{1/(n+4/6)−1/(n+1)}=π/3√3+1/3・log2

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