级数证明题

由f(x)以2π为周期, 有f(π) = f(-π).
代入分部积分公式可得:
b[n] = 1/π·∫{-π,π} f(x)sin(nx)dx
= 1/(nπ)·∫{-π,π} f'(x)cos(nx) dx (cos(nπ) = cos(-nπ))
= -1/(n^2·π)·∫{-π,π} f"(x)sin(nx) dx (sin(nπ) = sin(-nπ) = 0)
= -B[n]/n^2.
故2√(|b[n]|) = 2√(|B[n]|/n^2) ≤ 1/n^2+|B[n]| (均值不等式).
对n求和得2·∑{1 ≤ n} √(|b[n]|) ≤ ∑{1 ≤ n} 1/n^2 + ∑{1 ≤ n} |B[n]|.
再由∑{1 ≤ n} 1/n^2 = 1+∑{2 ≤ n} 1/n^2
< 1+∑{2 ≤ n} 1/(n(n-1))
= 1+∑{2 ≤ n} (1/(n-1)-1/n)
= 2,
即得∑{1 ≤ n} √(|b[n]|) < 1/2·(2+∑{1 ≤ n} |B[n]|).
由∑B[n]绝对收敛, 右端有界, 故左端正项级数∑{1 ≤ n} √(|b[n]|)也收敛.