证明：若{Un}满足Lim(n→∞)nUn=1,则∞∑(n=1) (-1)^n(Un+Un+1)收敛

其实只需试着写两项就能发现关键了.
那个级数写出来是-(U[1]+U[2])+(U[2]+U[3])-(U[3]+U[4])+...
除了U[1]以外的项都两两消掉了.
形式化的写出来是这样.
考虑级数∑{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].
由条件lim{n → ∞} n·U[n] = 1,有lim{n → ∞} U[n] = 0.
于是n → ∞时∑{1 ≤ k ≤ n} (-1)^k·(U[k]+U[k+1]) = -U[1]+(-1)^n·U[n+1]收敛到-U[1].
即级数∑{1 ≤ k} (-1)^k·(U[k]+U[k+1])收敛.