※ 引述《henry9621205 (清心小子)》之銘言：
: π π 2π nπ
: lim ----{cos---- + cos---- +........+cos---- }=?
: n→00 n 2n 2n 2n
π/n
--------- [sin(π/2n)][cos......]
sin(π/2n)
π/n
= --------- * 1/2 * {[sin(2π/2n) - sin0] + ....
sin(π/n)
..... [sin((n+1)π/2n) - sin((n-1)π/2n)]}
π/n
= ---------- * 1/2 * {sin[(n+1)π/2n] + sin[nπ/2n] - sinπ/2n}
sin(π/2n)
取極限
= 2
: 懇請解答 先謝過！！
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