19/07/17 00:03:47.43 d6A0g1cK.net
k≦x≦k+1 ⇒ 1/(k+1) ≦ 1/x ≦ 1/k,
ゆえ
I/(k+1) < ∫[k,k+1] |sin(πx)|/x dx < I/k,
ここに
I = ∫[k,k+1] |sin(πx)| dx = ∫[0,1] sin(πx) dx = [ -cos(πx)/π ](x=0,1) = 2/π,
(分子) > IΣ[k=1,n] 1/(k+1) > I∫[2,n+2] (1/x) dx = I {log(n+2) - log(2)},
(分子) < IΣ[k=1,n] 1/k < I{1 + ∫[1,n] (1/x) dx } = I {1 + log(n)},
∴ lim[n→∞] (分子)/log(n) = I = 2/π,