求助两道有关数列极限的小题1,如果ak（k是角标,下同）是单调的,且西格玛k=1到正无穷（ak）收敛,试证明lim k趋近正无穷 （k*ak）=02,如果ak是单调递减的,并且具有极限0,又对于一切k,bk=ak-2a（k+1

求助两道有关数列极限的小题1,如果ak（k是角标,下同）是单调的,且西格玛k=1到正无穷（ak）收敛,试证明lim k趋近正无穷 （k*ak）=02,如果ak是单调递减的,并且具有极限0,又对于一切k,bk=ak-2a（k+1
求助两道有关数列极限的小题
1,如果ak（k是角标,下同）是单调的,且西格玛k=1到正无穷（ak）收敛,试证明lim k趋近正无穷 （k*ak）=0
2,如果ak是单调递减的,并且具有极限0,又对于一切k,bk=ak-2a（k+1）+a（k+2）大于等于0【小括号里是角标】,试证明西格玛k=1到正无穷（k*bk）=a1
跪谢~

求助两道有关数列极限的小题1,如果ak（k是角标,下同）是单调的,且西格玛k=1到正无穷（ak）收敛,试证明lim k趋近正无穷 （k*ak）=02,如果ak是单调递减的,并且具有极限0,又对于一切k,bk=ak-2a（k+1
1. 不妨设a[k]单调递减, 由∑{1 ≤ k} a[k]收敛有lim{k → ∞} a[k] = 0, 于是a[k] ≥ 0.
仍由∑{1 ≤ k} a[k]收敛, 根据Cauchy收敛准则, 可得lim{n → ∞} ∑{n+1 ≤ k ≤ 2n} a[k] = 0.
而a[k]单调递减, 故a[k] ≥ a[2n]对k = n+1, n+2,..., 2n成立.
于是0 ≤ 2n·a[2n] ≤ 2·∑{n+1 ≤ k ≤ 2n} a[k] → 0 (n → ∞).
进而0 ≤ (2n+1)·a[2n+1] ≤ (2n+1)·a[2n] → 0 (n → ∞).
可知lim{k → ∞} k·a[k] = 0.
2. 用Abel求和公式.
∑{1 ≤ k ≤ n} k·b[k] = ∑{0 ≤ k ≤ n} k·b[k]
= ∑{1 ≤ k ≤ n} (k-(k-1))·∑{k ≤ j ≤ n} b[j]
= ∑{1 ≤ k ≤ n} ∑{k ≤ j ≤ n} b[j].
而∑{k ≤ j ≤ n} b[j] = ∑{k ≤ j ≤ n} ((a[j]-a[j+1])-(a[j+1]-a[j+2])) = (a[k]-a[k+1])-(a[n+1]-a[n+2]).
代入得∑{1 ≤ k ≤ n} k·b[k]
= ∑{1 ≤ k ≤ n} ((a[k]-a[k+1])-(a[n+1]-a[n+2]))
= -n·(a[n+1]-a[n+2])+∑{1 ≤ k ≤ n} (a[k]-a[k+1])
= -(n+1)·(a[n+1]-a[n+2])+a[1]-a[n+1]
而0 ≤ b[k] = (a[k]-a[k+1])-(a[k+1]-a[k+2]), 即数列a[k]-a[k+1]单调递减.
又∑{1 ≤ k ≤ n} (a[k]-a[k+1]) = a[1]-a[n+1] → a[1] (n → ∞).
由第1题的结论有(n+1)·(a[n+1]-a[n+2]) → 0 (n → ∞).
故∑{1 ≤ k ≤ n} k·b[k] = -(n+1)·(a[n+1]-a[n+2])+a[1]-a[n+1] → a[1] (n → ∞) (a[n+1]也收敛到0).

2、实在不好想明白的话就用归纳法。先证明最小的一项成立。然后假设第n项成立，证明弟(n+1)项成立。这样所有的都成立了。

1. Since a_k is mononic and the sum is convergent, we deduce |a_k| decreases and tends to 0. Thus we can suppose a_k>0 for all k>=1 without loss of generality.
Suppose claim is false. Then there ...

全部展开

1. Since a_k is mononic and the sum is convergent, we deduce |a_k| decreases and tends to 0. Thus we can suppose a_k>0 for all k>=1 without loss of generality.
Suppose claim is false. Then there exists epsilon>0 such that for any positive integer i, there exists k_i>i such that ka_k>epsilon. Since there is an infinite number of k_i>i such that k_ia_{k_i}>epsilon, we can take k_2>2k_1, k_3>2k_2, .... Consequently, by the mononicity of a_k we have
\sum_{i=1}^\infty a_k
>(a_1+...+a_{k_1})+(a_{k_1+1}+...+a_{k_2})+...
>epsilon [(1/k_1+...+1/k_1)+(1/k_2+...+1/k_2)+...]
=epsilon [1+(k_2-k_1)/k_2+(k_3-k_2)/k_3+...]
>epsilon [1+1/2+1/2+...].
This contradicts the hypothesis that \sum_{k=1}^\infty a_k is convergent. So \lim_{k=1}^\infty ka_k=0. Done.

收起