Posted by haifeng on 2025-01-18 10:34:34 last update 2025-01-18 10:34:34 | Edit | Answers (1)
设 xn=∑k=1n1k−2n, 证明 limn→∞xn 存在.
1
Posted by haifeng on 2025-01-18 12:47:41
证明:
(1) {xn} 是递减数列.
xn+1−xn=(∑k=1n+11k−2n+1)−(∑k=1n1k−2n)=1n+1−2(n+1−n)=1n+1−2n+1+n<0
(2) {xn} 存在下界.
利用
2(k+1−k)=2k+1+k<1k,
我们有
xn=∑k=1n1k−2n>∑k=1n2(k+1−k)−2n>2(n+1−1)−2n=2n+1+n−2>−2
因此, {xn} 有下界.
