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Question Number 59438 by tanmay last updated on 10/May/19
Commented by maxmathsup by imad last updated on 10/May/19
wehaveIn=πx=t∫0πtnπnsin(t)dtπ=1πn+1∫0πtnsintdtletfindWn=∫0πtnsintdtbypartsu′=tnandv=sint⇒Wn=[1n+1tn+1sint]0π−∫0πtn+1n+1costdt=−1n+1{[1n+2tn+2cost]0π+∫0πtn+2n+2sint}=−1n+1{1n+2(−πn+2)+1n+2∫0πtn+2sintdt}=πn+2(n+1)(n+2)−1(n+1)(n+2)∫0πtn+2sintdt⇒Wn=πn+2(n+1)(n+2)−1(n+1)(n+2)Wn+2wehaveIn=1πn+1Wn⇒In=π(n+1)(n+2)−1πn+1(n+1)(n+2)Wn+2=π(n+1)(n+2)−1πn+1(n+1)(n+2)πn+3In+2⇒In=π(n+1)(n+2)−π2(n+1)(n+2)In+2=π(n+1)(n+2){1−πIn+2}⇒In1−πIn+2=π(n+1)(n+2)⇒∑n=0∞In1−πIn+2=π∑n=0∞1(n+1)(n+2)=π∑n=1∞1n(n+1)letSn=∑k=1n1k(k+1)⇒Sn=∑k=1n(1k−1k+1)=1−12+12−13+....1n−1n+1=1−1n+1→1(n→+∞)⇒∑n=0∞{In1−πIn+2}=π.
Commented by tanmay last updated on 10/May/19
excellentsir...
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