■ある無限級数(その145)
(その140)〜(その144)をまとめておきたい.
[1]2の倍数の項のない交代級数
{1/1−1/3}+{1/3−1/5}+{1/5−1/7}+・・・
=Σ{1/(2k−1)−1/(2k+1)}=1
[2]3の倍数の項のない交代級数
{1/1−1/2+1/4−1/5}+{1/4−1/5+1/7−1/8}}+・・・
=Σ{1/(3k−2)−1/(3k−1)+1/(3k+1)−1/(3k+2)}
=1/2−(π/3)/tan(2π/3)
−1+(π/3)/tan(π/3)
[3]4の倍数の項のない交代級数
{1/1−1/2+1/3−1/5+1/6−1/7}+{1/5−1/6+1/7−1/9+1/10−1/12}}+・・・
=Σ{1/(4k−3)−1/(4k−2)+1/(4k−1)−1/(4k+1)+1/(4k+2)−1/(4k+3)}
=1/3−π/4/tan(3π/4)
−1/2
+1−π/4/tan(π/4)
[4]5の倍数の項のない交代級数
{1/1−1/2+1/3−1/4+1/6−1/7+1/8−1/9}+{1/6−1/7+1/8−1/9+1/11−1/12+1/13−1/14}}+・・・
=Σ{1/(5k−4)−1/(5k−3)+1/(5k−2)−1/(5k−1)+1/(5k+1)−1/(5k+2)+1/(5k+3)−1/(5k+4)}
=+1/4−π/5/tan(4π/5)
−1/3+π/5/tan(3π/5)
+1/2−(π/5)/tan(2π/5)
−1+π/5/tan(π/5)
[5]6の倍数の項のない交代級数
{1/1−1/2+1/3−1/4+1/5−1/7+1/8−1/9+1/10−1/12}+{1/7−1/8+1/9−1/10+1/11−1/13+1/14−1/15+1/16−1/17}}+・・・
=Σ{1/(6k−5)−1/(6k−4)+1/(6k−3)−1/(6k−2)+1/(6k−1)−1/(6k+1)+1/(6k+2)−1/(6k+3)+1/(6k+4)−1/(6k+5)}
=+1/5−π/6/tan(5π/6)
−1/4+π/6/tan(4π/6)
+1/3−(π/5)/tan(3π/6)
−1/2+π/6/tan(2π/6)
+1−π/6/tan(2π/6)
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[雑感]これではおもしろい形になっていない.
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