- 22 名前：１３２人目の素数さん mailto:sage [2009/06/20(土) 14:32:02 ]
- >>19
π/2 = p とおくと、相乗･相加平均より
(右辺) < ∑[k=1,n-1] (2/3k){log(kp) + log((k+1)p)}
= ∑[k=1,n-1] (2/3k){log(k) + log(k+1) + 2log(p)},
(左辺) = log(n)log(np^2) = log(n){log(n) + 2log(p)} = log(n)^2 + 2log(p)･log(n),
したがって、
∑[k=1,n-1] (2/3k){log(k) + log(k+1)} < log(n)^2, ･････ (I)
∑[k=1,n-1] (2/3k) < log(n), ･････ (II)
を示せば十分。
(I)
1/(k+1) < -log(k/(k+1)) = log((k+1)/k) = log(k+1) - log(k),
より
∑[k=1,n-1] (2/3k){log(k+1) + log(k)} < (2/3)log(2) + ∑[k=2,n-1] (1/(k+1)){log(k+1) + log(k)}
= (2/3)log(2) + ∑[k=2,n-1] {log(k+1) - log(k)}{log(k+1) + log(k)}
= (2/3)log(2) + ∑[k=2,n-1] {log(k+1)^2 - log(k)^2}
= (2/3)log(2) + log(n)^2 - log(2)^2
= log(n)^2 -log(2){log(2) -2/3}
< log(n)^2, {log(2) = 0.693147･･･ >2/3}
(II)
・n=2 のときは 明らか。
・n>2 のとき、(I) と同様に
∑[k=1,n-1] (2/3k) = 2/3 + ∑[k=2,n-1] (2/3k)
< 2/3 + ∑[k=2,n-1] {log(k+1)-log(k)}
= 2/3 + log(n) - log(2)
< log(n), {log(2) = 0.693147･･･ >2/3}
または、 y=1/x が下に凸だから
∑[k=1,n-1] (1/k) < ∫[1/2, n -1/2] 1/x dx = log(2n-1) < (3/2)log(n),
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