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Question Number 121498 by Lordose last updated on 08/Nov/20
Evaluate Σ_(n=1) ^∞ ((sin(n))/n)
Evaluate∑∞n=1sin(n)n
Answered by Bird last updated on 09/Nov/20
let find S(x) =Σ_(n=1) ^(∞ ) ((sin(nx))/n)dx S(x)=Im(Σ_(n=1) ^∞ (e^(inx) /n)) and Σ_(n=1) ^(∞ ) (e^(inx) /n) =Σ_(n=1) ^(∞ ) (((e^(ix) )^n )/n) =−ln(1−e^(ix) ) =−ln(1−cosx−isinx ) =−ln(2sin^2 ((x/2))−2isin((x/2))cos((x/2))) =−ln(−2isin((x/2))e^((ix)/2) ) =−ln(−2)−ln(i)−ln(sin((x/2)))−((ix)/2) =−ln(2)−iπ−((iπ)/2)−((ix)/2)−ln(sin((x/2))) =−ln2−ln(sin((x/2)))−((ix)/2)−((3iπ)/2) ⇒Σ ((sin(nx))/n) =−((x+3π)/2) x=1 ⇒ Σ ((sin(n))/n) =−((3π+1)/2)
letfindS(x)=∑n=1∞sin(nx)ndxS(x)=Im(∑n=1∞einxn)and∑n=1∞einxn=∑n=1∞(eix)nn=−ln(1−eix)=−ln(1−cosx−isinx)=−ln(2sin2(x2)−2isin(x2)cos(x2))=−ln(−2isin(x2)eix2)=−ln(−2)−ln(i)−ln(sin(x2))−ix2=−ln(2)−iπ−iπ2−ix2−ln(sin(x2))=−ln2−ln(sin(x2))−ix2−3iπ2⇒Σsin(nx)n=−x+3π2x=1⇒Σsin(n)n=−3π+12

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