※ 引述《PttFund (批踢踢基金只進不出)》之銘言:
: 決定下列級數, 對於實數 x, 何時收斂?
: ∞ k sin(kx)
: Σ ( Σ 1/s)(---------).
: k=1 s=1 k
k
令 a_k = (Σ 1/s)/k
s=1
Claim : (1) a_k 遞減至 0 當 k -> oo
n
(2) {Σ sin(kx) : n in N} 有界 whenever x in [δ, 2π-δ], (0<δ<π)
k=1
k k+1 k k
(Σ 1/s)/k - (Σ 1/s)/(k+1) = [(k+1)(Σ1/s) - k(Σ1/s) - k/(k+1)]/k(k+1)
s=1 s=1 s=1 s=1
k
= [(Σ1/s) - k/(k+1)]/k(k+1) > [1 - k/(k+1)]/k(k+1) > 0 for k ≧ 1
s=1
故 a_k 遞減且因為 1/k -> 0 as k -> oo
故極限的平均值定理 => a_k -> 0 as k -> oo
n
接著令 S(n,x) = Σsin(kx), 在等號兩邊同乘 2sin(x/2) 積化和差推得
k=1
n n
2sin(x/2)S(n,x) = Σ 2sin(x/2)sin(kx) = Σ {cos[(k-1/2)x] - cos[(k+1/2)x]}
k=1 k=1
= cos(x/2) - cos[(n+1/2)x] = 2sin(nx/2)sin[(n+1)x/2]
故 |S(n,x)| = |sin(nx/2)sin[(n+1)x/2]/sin(x/2)|
≦ |csc(x/2)| ≦ |csc(δ/2)| for all n and x
故 Claim (2) 成立
Dirichlet test => 原級數收斂 whenever x in [δ, 2π-δ], (0<δ<π)
事實上當 x 在 [2nπ+δ, 2nπ+2π-δ] 上時原級數都收斂, n in Z
又當 x = 2nπ 時原級數也收斂
故原級數對所有 x in R 收斂
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