■素数と無限級数(その141)
四角数の逆数和
Σ1/n^2 =1/1^2+1/2^2+1/3^2+1/4^2+・・・
はπ^2/6に収束する.
===================================
多角数の逆数和は
[1]三角数:n(n+1)/2→Σ2/n(n+1)=2
[2]四角数:n^2→Σ1/n^2=π^2/6
[3]五角数:(3n^2−n)/2→Σ2/n(3n−1)=3log3−π/√3
Σ2/n(3n−1) (n=1〜)
=Σ2/(n+1)(3n+2) (n=0〜)
=6Σ1/(n+1)(n+2/3) (n=0〜)
=2Σ{1/(n+2/3)−1/(n+1)} (n=0〜)
Σ{1/(n+2/3)−1/(n+1)}=−π/2√3+log12−2{−1/2・log1/2−1/2・log√3/2}
=−π/2√3+2log2+log3−log2+(1/2・log3−log2)
=−π/2√3+3/2・log3
したがって,
Σ2/n(3n−1)=3log3−π/√3
===================================
[4]六角数:2n^2−n→Σ1/2n(2n−1)=2log2
[5]七角数:n(5n−3)/2→Σ2/n(5n−3)=π/3・(1−2/√5)^1/2+5/6・log5−√5/3・log(φ)
[6]八角数:n(6n−4)/2→Σ2/n(6n−4)π/4√3+3/4・log3
===================================