■逆数和とその積分表示(その8)

  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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