the options are: 0, infinity, n , and n(n+1)/2.
lim x->∞ {11/sin2x + 21/sin2x + .... + n1/sin2x} sin2x
??????
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4 Answers
ith_power
·2008-11-24 06:08:24
given=
(i=0nΣ(i/n)^(1/sin2x))sin2x* n= 0 , since 1/sin2x≥1 and i/n<=1
what's the answer??
ith_power
·2008-11-24 06:25:14
oops sorry, one term inside was n/n, it will give 1 not 0. so answer=1.n=n