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Question Number 124355 by Eric002 last updated on 02/Dec/20
findthelimitlimn→∞(sin(1n2)+sin(2n2)+......+sin(nn2))
Answered by Dwaipayan Shikari last updated on 02/Dec/20
limn→∞sin(1n2)+sin(2n2)+...=1+2+3+4+5+6+..+nn2=n2+n2n2=12(Assin(1n2)→(1n2))
Answered by mathmax by abdo last updated on 02/Dec/20
wehavex−x36⩽sinx⩽x⇒∑k=1nkn2−16∑k=1nk3n6⩽∑k=1nsin(kn2)⩽∑k=1nkn2⇒n(n+1)2n2−16n3(n(n+1)2)2⩽Sn⩽n(n+1)2n2wepasseto[limit(n→∞)limn→+∞Sn=12
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