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find-lim-n-1-n-2-sin-1-n-2-2sin-4-n-2-n-1-sin-n-1-2-n-2-




Question Number 66171 by mathmax by abdo last updated on 10/Aug/19
find lim_(n→+∞)  (1/n^2 ){sin((1/n^2 ))+2sin((4/n^2 ))+....(n−1)sin((((n−1)^2 )/n^2 ))}
findlimn+1n2{sin(1n2)+2sin(4n2)+.(n1)sin((n1)2n2)}
Commented by mathmax by abdo last updated on 10/Aug/19
let S_n =(1/n^2 ){sin((1/n^2 ))+2sin((4/n^2 ))+...+(n−1)sin((((n−1)^2 )/n^2 ))}  we have S_n =(1/n^2 )Σ_(k=0) ^(n−1)  k sin((k^2 /n^2 )) =(1/n)Σ_(k=0) ^(n−1) ((k/n))sin((k^2 /n^2 ))  S_n is a Rieman sum ⇒lim_(n→+∞)  S_n =∫_0 ^1 xsin(x^2 )dx  =[−(1/2)cos(x^2 )]_0 ^1  =−(1/2)(cos(1)−1)=(1/2) −((cos(1))/2).
letSn=1n2{sin(1n2)+2sin(4n2)++(n1)sin((n1)2n2)}wehaveSn=1n2k=0n1ksin(k2n2)=1nk=0n1(kn)sin(k2n2)SnisaRiemansumlimn+Sn=01xsin(x2)dx=[12cos(x2)]01=12(cos(1)1)=12cos(1)2.
Answered by Smail last updated on 10/Aug/19
lim_(n→∞) (1/n^2 )(sin(1/n^2 )+2sin(4/n^2 )+...+(n−1)sin((((n−1)^2 )/n^2 )))  lim_(n→∞) (1/n)×(1/n)(sin(1/n^2 )+2sin(4/n^2 )+...+(n−1)sin((((n−1)^2 )/n^2 )))  lim_(n→∞) (1/n)((1/n)sin(1/n^2 )+(2/n)sin(4/n^2 )+...+((n−1)/n)sin((((n−1)^2 )/n^2 )))  =lim_(n→∞) ((1−0)/n)Σ_(k=1) ^(n−1) (k/n)sin(((k/n))^2 )  =∫_0 ^1 xsin(x^2 )dx−lim_(n→∞) (1/n)((n/n)sin(((n/n))^2 ))  =−(1/2)[cos(x^2 )]_0 ^1 =(1/2)(1−cos(1))  =sin^2 ((1/2))
limn1n2(sin(1/n2)+2sin(4/n2)++(n1)sin((n1)2n2))limn1n×1n(sin(1/n2)+2sin(4/n2)++(n1)sin((n1)2n2))limn1n(1nsin(1/n2)+2nsin(4/n2)++n1nsin((n1)2n2))=limn10nn1k=1knsin((kn)2)=01xsin(x2)dxlimn1n(nnsin((nn)2))=12[cos(x2)]01=12(1cos(1))=sin2(12)

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