find-lim-n-1-n-2-sin-1-n-2-2sin-4-n-2-n-1-sin-n-1-2-n-2- Tinku Tara June 3, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 66171 by mathmax by abdo last updated on 10/Aug/19 findlimn→+∞1n2{sin(1n2)+2sin(4n2)+….(n−1)sin((n−1)2n2)} Commented by mathmax by abdo last updated on 10/Aug/19 letSn=1n2{sin(1n2)+2sin(4n2)+…+(n−1)sin((n−1)2n2)}wehaveSn=1n2∑k=0n−1ksin(k2n2)=1n∑k=0n−1(kn)sin(k2n2)SnisaRiemansum⇒limn→+∞Sn=∫01xsin(x2)dx=[−12cos(x2)]01=−12(cos(1)−1)=12−cos(1)2. Answered by Smail last updated on 10/Aug/19 limn→∞1n2(sin(1/n2)+2sin(4/n2)+…+(n−1)sin((n−1)2n2))limn→∞1n×1n(sin(1/n2)+2sin(4/n2)+…+(n−1)sin((n−1)2n2))limn→∞1n(1nsin(1/n2)+2nsin(4/n2)+…+n−1nsin((n−1)2n2))=limn→∞1−0n∑n−1k=1knsin((kn)2)=∫01xsin(x2)dx−limn→∞1n(nnsin((nn)2))=−12[cos(x2)]01=12(1−cos(1))=sin2(12) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: prove-without-calculus-that-0-cos-x-2-dx-0-sin-x-2-dx-Next Next post: x-x-x-x-2-x- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.