其實只需試著寫兩項就能發(fā)現(xiàn)關(guān)鍵了.
那個級數(shù)寫出來是-(U[1]+U[2])+(U[2]+U[3])-(U[3]+U[4])+...
除了U[1]以外的項都兩兩消掉了.
形式化的寫出來是這樣.
考慮級數(shù)∑{1 ≤ k} (-1)^k·(U[k]+U[k+1])的部分和:
∑{1 ≤ k ≤ n} (-1)^k·(U[k]+U[k+1])
= ∑{1 ≤ k ≤ n} (-1)^k·U[k]+∑{1 ≤ k ≤ n} (-1)^k·U[k+1]
= -U[1]+∑{2 ≤ k ≤ n} (-1)^k·U[k]+(-1)^n·U[n+1]+∑{1 ≤ k ≤ n-1} (-1)^k·U[k+1]
= -U[1]+(-1)^n·U[n+1]+∑{2 ≤ k ≤ n} (-1)^k·U[k]+∑{2 ≤ k ≤ n} (-1)^(k-1)·U[k]
= -U[1]+(-1)^n·U[n+1]+∑{2 ≤ k ≤ n} (-1)^k·U[k]-∑{2 ≤ k ≤ n} (-1)^k·U[k]
= -U[1]+(-1)^n·U[n+1].
由條件lim{n → ∞} n·U[n] = 1,有l(wèi)im{n → ∞} U[n] = 0.
于是n → ∞時∑{1 ≤ k ≤ n} (-1)^k·(U[k]+U[k+1]) = -U[1]+(-1)^n·U[n+1]收斂到-U[1].
即級數(shù)∑{1 ≤ k} (-1)^k·(U[k]+U[k+1])收斂.