\(\displaystyle {\frac {\pi }{2}}=\prod {n=1}^{\infty }{\frac {4n^{2}} {4n^{2}-1}}=\prod {n=1}^{\infty }{\frac {(2n)^{2}}{(2n-1)(2n+1)}}\)
RSS Link – 雑記帳だけ
Theme laks By icysamon|プライバシーポリシー
Copyright © 2023 – 2025 icysamon