2个回答
展开全部
(3)
(1)
(n+1)/(2n)^2 <∑(i:0->n) 1/(n+i)^2 <(n+1)/n^2
lim(n->∞) (n+1)/(2n)^2 = lim(n->∞) (n+1)/n^2=0
=>
lim(n->∞) ∑(i:0->n) 1/(n+i)^2 =0
(2)
2^n/n!>0
2^n/n!
=(2/1)(2/2)[ (2/3)(2/4)...(2/n)]
=2[ (2/3)(2/4)...(2/n)]
<2(2/3)^(n-2)
lim(n->∞) 2(2/3)^(n-2) =0
=>
lim(n->∞) 2^n/n! =0
(1)
(n+1)/(2n)^2 <∑(i:0->n) 1/(n+i)^2 <(n+1)/n^2
lim(n->∞) (n+1)/(2n)^2 = lim(n->∞) (n+1)/n^2=0
=>
lim(n->∞) ∑(i:0->n) 1/(n+i)^2 =0
(2)
2^n/n!>0
2^n/n!
=(2/1)(2/2)[ (2/3)(2/4)...(2/n)]
=2[ (2/3)(2/4)...(2/n)]
<2(2/3)^(n-2)
lim(n->∞) 2(2/3)^(n-2) =0
=>
lim(n->∞) 2^n/n! =0
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展开全部
1/n^2 + 1/(n+1)^2+ ... + 1/(n+n)^2 > 1/(n+n)^2+1/(n+n)^2 + ... + 1/(n+n)^2 = n/(2n)^2 ->0
1/n^2 + 1/(n+1)^2+ ... + 1/(n+n)^2 < 1/n^2 + 1/n^2 +...+1/n^2 = n/n^2 ->0
得证
1/n^2 + 1/(n+1)^2+ ... + 1/(n+n)^2 < 1/n^2 + 1/n^2 +...+1/n^2 = n/n^2 ->0
得证
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