■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)

===================================

[雑感]これではおもしろい形になっていない.

===================================