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
は
1/2・Σ{1/(n+1/2)-1/(n+1)}
1-1/3+1/5-1/7+・・・=π/4
は
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+・・・=?
は
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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