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Question Number 31707 by gunawan last updated on 13/Mar/18
Findlimx→∞Σnk=11nsin2πn
Commented by abdo imad last updated on 18/Mar/18
ithinktheQ.isfindlimn→∞∑k=1n1nsin(2kπn)letputSn=∑k=1n1nsin(2kπn)wehaveSn=12π2πn∑k=1nsin(k2πn)→12π∫02πsinxdx=12π[−cosx]02π=0⇒limn→∞Sn=0
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