Question

计算下列数列极限？

利用等差数列 等比数列求和公式和夹逼准则

Answer 1

很简单的，看到根号首先想到有理化

Answer 2

(6)
f(x)=e^x
lim(n->∞) (1/n^2) ln[f(1)f(2)...f(n) ]
=lim(n->∞) (1/n^2)∑(i:1->n) i
=lim(n->∞) (1/n^2) [n(n+1)/2]
=1/2
(7)
lim(n->∞) [√(1+2+..+n) -√(1+2+..+(n-1))]
=lim(n->∞) [(1+2+..+n) -(1+2+..+(n-1))]/[√(1+2+..+n) +√(1+2+..+(n-1))]
=lim(n->∞) n/[√(1+2+..+n) +√(1+2+..+(n-1))]
=lim(n->∞) n/{ √[n(n+1)/2] +√[(n-1)n/2] }
分子分母同时除以n
=lim(n->∞) 1/{ √[(1+1/n)/2] +√[(1-1/n)/2] }
=1/[2√(1/2)]
=√2/2
(8)
2012. (1)^(1/n)<1^(1/n) +2^(1/n) +...+2012^(1/n) < 2012.(2012)^(1/n)
lim(n->∞) 2012. (1)^(1/n) =lim(n->∞) 2012. (2012)^(1/n) =2012
=> lim(n->∞) [1^(1/n) +2^(1/n) +...+2012^(1/n) ] =2012

(9)
lim(n->∞) [√(n+√n) -√(n-√n)]
=lim(n->∞) [(n+√n) -(n-√n)] /[√(n+√n) +√(n-√n)]
=lim(n->∞) 2√n /[√(n+√n) +√(n-√n)]
分子分母同时除以√n
=lim(n->∞) 21 /[√(1+1/√n) +√(1-1/√n)]
=2/(1+1)
=1
(10)
n(n+1)/[2(n^2-n-n)] <1/(n^2-n-1) +2/(n^2-n-2) +...+n/(n^2+n-n) and
n(n+1)/[2(n^2-n-1)] > 1/(n^2-n-1) +2/(n^2-n-2) +...+n/(n^2+n-n)
lim(n->∞)n(n+1)/[2(n^2-n-n)] = lim(n->∞)n(n+1)/[2(n^2-n-1)] = 1/2
=>
lim(n->∞) [1/(n^2-n-1) +2/(n^2-n-2) +...+n/(n^2+n-n)] =1/2

Answer 3

详见下图，望采纳。