Evaluating $sum_{n=1}^infty frac{sin(nx)}{n}$ without integrating $sum_{n=1}^infty e^{nx}$












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I am looking for alternative solutions for finding this sum
$$sum_{n=1}^infty frac{sin(nx)}{n} $$




My solution proceeds by integrating $$sum_{n=1}^infty e^{nx}=frac{e^{ix}}{1-e^{ix}}$$
With suitable limits and then taking the imaginary part of it.






integration sequences-and-series trigonometry summation







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    2












    $begingroup$


    I am looking for alternative solutions for finding this sum
    $$sum_{n=1}^infty frac{sin(nx)}{n} $$




    My solution proceeds by integrating $$sum_{n=1}^infty e^{nx}=frac{e^{ix}}{1-e^{ix}}$$
    With suitable limits and then taking the imaginary part of it.






    integration sequences-and-series trigonometry summation







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I am looking for alternative solutions for finding this sum
      $$sum_{n=1}^infty frac{sin(nx)}{n} $$




      My solution proceeds by integrating $$sum_{n=1}^infty e^{nx}=frac{e^{ix}}{1-e^{ix}}$$
      With suitable limits and then taking the imaginary part of it.






      integration sequences-and-series trigonometry summation







      share|cite|improve this question











      $endgroup$




      I am looking for alternative solutions for finding this sum
      $$sum_{n=1}^infty frac{sin(nx)}{n} $$




      My solution proceeds by integrating $$sum_{n=1}^infty e^{nx}=frac{e^{ix}}{1-e^{ix}}$$
      With suitable limits and then taking the imaginary part of it.






      integration sequences-and-series trigonometry summation




      integration sequences-and-series trigonometry summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 24 at 12:41









      Blue

      49.1k870156




      49.1k870156










      asked Jan 24 at 11:03









      KarthikKarthik

      13212




      13212






















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          Note thatbegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=operatorname{Im}left(sum_{n=1}^inftyfrac{e^{inx}}nright)\&=operatorname{Im}left(sum_{n=1}^inftyfrac{left(e^{ix}right)^n}nright).end{align}If $xinmathbb{R}setminus{0}$, then $e^{ix}in S^1setminus{1}$, and therefore$$sum_{n=1}^inftyfrac{left(e^{ix}right)^n}n=logleft(frac1{1-e^{ix}}right),$$(where $log$ is the principal branch of the logarithm), sincebegin{align}log(1+x)=x-frac{x^2}2+frac{x^3}3-frac{x^4}x+cdots&implieslog(1-x)=-left(x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsright)\&implieslogleft(frac1{1-x}right)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsend{align}Thereforebegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=-arctanleft(frac{-sin x}{1-cos(x)}right)\&=arctanleft(cotleft(frac x2right)right).end{align}Of course, if $x=0$, then this equality also holds.






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            $begingroup$

            Note thatbegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=operatorname{Im}left(sum_{n=1}^inftyfrac{e^{inx}}nright)\&=operatorname{Im}left(sum_{n=1}^inftyfrac{left(e^{ix}right)^n}nright).end{align}If $xinmathbb{R}setminus{0}$, then $e^{ix}in S^1setminus{1}$, and therefore$$sum_{n=1}^inftyfrac{left(e^{ix}right)^n}n=logleft(frac1{1-e^{ix}}right),$$(where $log$ is the principal branch of the logarithm), sincebegin{align}log(1+x)=x-frac{x^2}2+frac{x^3}3-frac{x^4}x+cdots&implieslog(1-x)=-left(x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsright)\&implieslogleft(frac1{1-x}right)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsend{align}Thereforebegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=-arctanleft(frac{-sin x}{1-cos(x)}right)\&=arctanleft(cotleft(frac x2right)right).end{align}Of course, if $x=0$, then this equality also holds.






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              $begingroup$

              Note thatbegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=operatorname{Im}left(sum_{n=1}^inftyfrac{e^{inx}}nright)\&=operatorname{Im}left(sum_{n=1}^inftyfrac{left(e^{ix}right)^n}nright).end{align}If $xinmathbb{R}setminus{0}$, then $e^{ix}in S^1setminus{1}$, and therefore$$sum_{n=1}^inftyfrac{left(e^{ix}right)^n}n=logleft(frac1{1-e^{ix}}right),$$(where $log$ is the principal branch of the logarithm), sincebegin{align}log(1+x)=x-frac{x^2}2+frac{x^3}3-frac{x^4}x+cdots&implieslog(1-x)=-left(x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsright)\&implieslogleft(frac1{1-x}right)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsend{align}Thereforebegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=-arctanleft(frac{-sin x}{1-cos(x)}right)\&=arctanleft(cotleft(frac x2right)right).end{align}Of course, if $x=0$, then this equality also holds.






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                7












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                7





                $begingroup$

                Note thatbegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=operatorname{Im}left(sum_{n=1}^inftyfrac{e^{inx}}nright)\&=operatorname{Im}left(sum_{n=1}^inftyfrac{left(e^{ix}right)^n}nright).end{align}If $xinmathbb{R}setminus{0}$, then $e^{ix}in S^1setminus{1}$, and therefore$$sum_{n=1}^inftyfrac{left(e^{ix}right)^n}n=logleft(frac1{1-e^{ix}}right),$$(where $log$ is the principal branch of the logarithm), sincebegin{align}log(1+x)=x-frac{x^2}2+frac{x^3}3-frac{x^4}x+cdots&implieslog(1-x)=-left(x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsright)\&implieslogleft(frac1{1-x}right)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsend{align}Thereforebegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=-arctanleft(frac{-sin x}{1-cos(x)}right)\&=arctanleft(cotleft(frac x2right)right).end{align}Of course, if $x=0$, then this equality also holds.






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                Note thatbegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=operatorname{Im}left(sum_{n=1}^inftyfrac{e^{inx}}nright)\&=operatorname{Im}left(sum_{n=1}^inftyfrac{left(e^{ix}right)^n}nright).end{align}If $xinmathbb{R}setminus{0}$, then $e^{ix}in S^1setminus{1}$, and therefore$$sum_{n=1}^inftyfrac{left(e^{ix}right)^n}n=logleft(frac1{1-e^{ix}}right),$$(where $log$ is the principal branch of the logarithm), sincebegin{align}log(1+x)=x-frac{x^2}2+frac{x^3}3-frac{x^4}x+cdots&implieslog(1-x)=-left(x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsright)\&implieslogleft(frac1{1-x}right)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+cdotsend{align}Thereforebegin{align}sum_{n=1}^inftyfrac{sin(nx)}n&=-arctanleft(frac{-sin x}{1-cos(x)}right)\&=arctanleft(cotleft(frac x2right)right).end{align}Of course, if $x=0$, then this equality also holds.







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                edited Jan 25 at 13:35

























                answered Jan 24 at 11:11









                José Carlos SantosJosé Carlos Santos

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