Question :
http://smcc.edu.hk/math/f67/exercises/Infinite%20Sequence%20(exercise).pdf
[Question 11(b)] --- I don't understand how we can assume that [(n+1)1/2n<...... (n+1)1/n]
Solution :
http://smcc.edu.hk/math/f67/exercises%20solution/Infinite%20Sequences%20(solution).pdf
Pure maths --- Infinite Sequences
2007-06-24 7:15 am
回答 (2)
2007-06-24 7:46 am
參考: My Maths knowledge
2007-06-24 7:40 am
First,
1/(n+k) < 1/n for all k = 1,2, ..., n.
Also,
1/(n+k) > 1/(2n) for all k = 0,1,..., n-1.
So, we have
1/n + 1/(n+1) + 1/(n+2) +...+ 1/(2n) < 1/n + 1/n + 1/n + ... + 1/n = (n+1) 1/n.
Similary, we have
1/n + 1/(n+1) + 1/(n+2) +...+ 1/(2n) > 1/(2n) + 1/(2n) + 1/(2n) + ... + 1/(2n) = (n+1) 1/(2n).
1/(n+k) < 1/n for all k = 1,2, ..., n.
Also,
1/(n+k) > 1/(2n) for all k = 0,1,..., n-1.
So, we have
1/n + 1/(n+1) + 1/(n+2) +...+ 1/(2n) < 1/n + 1/n + 1/n + ... + 1/n = (n+1) 1/n.
Similary, we have
1/n + 1/(n+1) + 1/(n+2) +...+ 1/(2n) > 1/(2n) + 1/(2n) + 1/(2n) + ... + 1/(2n) = (n+1) 1/(2n).
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070623000051KK04849