■ある無限級数(その172)

  1−1/2+1/3−1/4+・・・=log2

  1−1/3+1/5−1/7+・・・=π/4

  1−1/4+1/7−1/10+・・・=π/3√3+1/3・log2

 それでは,

[Q]1−1/5+1/9−1/13+・・・=?

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  1−1/5+1/9−1/13+・・・=?

  1/8・Σ{1/(n+1/8)−1/(n+1)}

 −1/8・Σ{1/(n+5/8)−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=8

  Σ{1/(n+p/q)−1/(n+1)}=π/2・(√2+1)+log16−log16

[2]p=5,q=8

  Σ{1/(n+p/q)−1/(n+1)}=−π/2・(√2+1)+log16−2{log(2+√2)}

  1/8・Σ{1/(n+1/8)−1/(n+1)}

 −1/8・Σ{1/(n+5/8)−1/(n+1)}

 ={π(√2+1)-log16+2{log(2+√2)}}/8

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