use integration to evaluate the limit:
lim(n->infinity) {[1/[root(n)*root(n+1)]+[1/[root(n)*root(n+2)]+...+[1/[root(n)*root(n+n)]
(use appropriate Riemann Sum to help)
Thanks..
integration to evaluate limit
2012-10-02 12:20 am
回答 (2)
2012-10-02 4:29 am
✔ 最佳答案
lim (n->∞) 1/√n(n + 1) + 1/√n(n + 2) + 1/√n(n + 3) + ... + 1/√n(n + n)= lim (n->∞) (1/n)√[n/(n + 1)] + (1/n)√[n/(n + 2)] + (1/n)√[n/(n + 3)] + ... + (1/n)√[n/(n + n)]
= lim (n->∞) (1/n)√[1/(1 + 1/n)] + (1/n)√[1/(1 + 2/n)] + (1/n)√[1/(1 + 3/n)] + ... + (1/n)√[1/(1 + n/n)]
= ∫ (0->1) 1/(1 + x) dx
= ln |1 + x| |[0,1]
= ln 2
2012-10-02 20:52:46 補充：
Forth line should be
= ∫ (0->1) 1/√(1 + x) dx
= 2√(1 + x) | [0,1]
= 2(√2 - 1)
2012-10-02 8:49 pm
lim (n->∞) (1/n)√[1/(1 + 1/n)] + (1/n)√[1/(1 + 2/n)] + (1/n)√[1/(1 + 3/n)] + ... + (1/n)√[1/(1 + n/n)]
= ∫ (0->1) 1/(1 + x) dx <----- any square root here?
= ∫ (0->1) 1/(1 + x) dx <----- any square root here?
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