Menu Close

Evaluate-n-1-sin-n-n-

Tinku Tara 0 Comments
Pin




Question Number 121498 by Lordose last updated on 08/Nov/20
Evaluate Σ_(n=1) ^∞ ((sin(n))/n)
$$ \\ $$$$\mathrm{Evaluate} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\mathrm{n}\right)}{\mathrm{n}} \\ $$
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)
$${let}\:{find}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{{sin}\left({nx}\right)}{{n}}{dx} \\ $$$${S}\left({x}\right)={Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right)\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{{e}^{{inx}} }{{n}}\:=\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}} \\ $$$$=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{cosx}−{isinx}\:\right) \\ $$$$=−{ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\left(−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){e}^{\frac{{ix}}{\mathrm{2}}} \right) \\ $$$$=−{ln}\left(−\mathrm{2}\right)−{ln}\left({i}\right)−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{ix}}{\mathrm{2}} \\ $$$$=−{ln}\left(\mathrm{2}\right)−{i}\pi−\frac{{i}\pi}{\mathrm{2}}−\frac{{ix}}{\mathrm{2}}−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\mathrm{2}−{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{ix}}{\mathrm{2}}−\frac{\mathrm{3}{i}\pi}{\mathrm{2}} \\ $$$$\Rightarrow\Sigma\:\frac{{sin}\left({nx}\right)}{{n}}\:=−\frac{{x}+\mathrm{3}\pi}{\mathrm{2}} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\Sigma\:\:\frac{{sin}\left({n}\right)}{{n}}\:=−\frac{\mathrm{3}\pi+\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *