sin⁡π3+12sin⁡2π3+13sin⁡3π3+……..∞=

Let S=sin⁡π3+12sin⁡2π3+13sin⁡3π3+……∞C=cos⁡π3+12cos⁡2π3+13cos⁡3π3+……∞c+is=cos⁡π3+isin⁡π3+12cos⁡2π3+isin⁡2π3+13cos⁡3π3+isin⁡3π3+……∞=eiπ3+12eiπ32+13eiπ33+………∞=−log⁡1−eiπ3=−log⁡1−cos⁡π3+isin⁡π3=−log⁡1−12−i32=−log⁡12−i32=−log⁡14+34+iArg⁡12−i32 (∵log⁡z=log⁡|z|+iArg⁡z)=−log⁡1+itan−1⁡(−3)=itan−1⁡3⇒S=π3

sin⁡π3+12sin⁡2π3+13sin⁡3π3+……..∞=

Let S=sin⁡π3+12sin⁡2π3+13sin⁡3π3+……∞C=cos⁡π3+12cos⁡2π3+13cos⁡3π3+……∞c+is=cos⁡π3+isin⁡π3+12cos⁡2π3+isin⁡2π3+13cos⁡3π3+isin⁡3π3+……∞=eiπ3+12eiπ32+13eiπ33+………∞=−log⁡1−eiπ3=−log⁡1−cos⁡π3+isin⁡π3=−log⁡1−12−i32=−log⁡12−i32=−log⁡14+34+iArg⁡12−i32 (∵log⁡z=log⁡|z|+iArg⁡z)=−log⁡1+itan−1⁡(−3)=itan−1⁡3⇒S=π3

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$\sin \frac{π}{3} + \frac{1}{2} \sin \frac{2 π}{3} + \frac{1}{3} \sin \frac{3 π}{3} + \dots \dots . . \infty =$

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$π$ /4

$π$ /3

$π$ /2

answer is D.

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Detailed Solution

$\begin{array}{l} Let S = \sin \frac{π}{3} + \frac{1}{2} \sin \frac{2 π}{3} + \frac{1}{3} \sin \frac{3 π}{3} + \dots \dots \infty \\ C = \cos \frac{π}{3} + \frac{1}{2} \cos \frac{2 π}{3} + \frac{1}{3} \cos \frac{3 π}{3} + \dots \dots \infty \\ c + is = (\cos \frac{π}{3} + isin \frac{π}{3}) + \frac{1}{2} (\cos \frac{2 π}{3} + isin \frac{2 π}{3}) + \frac{1}{3} (\cos \frac{3 π}{3} + isin \frac{3 π}{3}) + \dots \dots \infty \\ = e^{\frac{iπ}{3}} + \frac{1}{2} {(e^{\frac{iπ}{3}})}^{2} + \frac{1}{3} {(e^{\frac{iπ}{3}})}^{3} + \dots \dots \dots \infty \\ = - \log (1 - e^{\frac{iπ}{3}}) \\ = - \log (1 - (\cos \frac{π}{3} + isin \frac{π}{3})) \end{array}$
$\begin{array}{l} = - \log ((1 - \frac{1}{2}) - i \frac{\sqrt{3}}{2}) \\ = - \log (\frac{1}{2} - i \frac{\sqrt{3}}{2}) \\ = - (\log \sqrt{\frac{1}{4} + \frac{3}{4}} + iArg (\frac{1}{2} - i \frac{\sqrt{3}}{2})) (∵ \log z = \log | z | + iArg z) \\ = - (\log 1 + {itan}^{- 1} (- \sqrt{3})) \\ = {itan}^{- 1} \sqrt{3} \\ \Rightarrow S = \frac{π}{3} \end{array}$