Sin⁡ π3+12sin⁡ 2π3+13sin ⁡3π3+…∞=

Let S=sin⁡ π3+12sin ⁡2π3+13sin ⁡3π3+…∞C=cos⁡π3+12cos⁡2π3+13cos⁡3π3+…∞c+is=cis⁡π3+12cis⁡2π3+13cis⁡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|+iArgZ)=−log⁡1+itan−1⁡(−3)=itan−1⁡3=iπ3

Sin⁡ π3+12sin⁡ 2π3+13sin ⁡3π3+…∞=

Let S=sin⁡ π3+12sin ⁡2π3+13sin ⁡3π3+…∞C=cos⁡π3+12cos⁡2π3+13cos⁡3π3+…∞c+is=cis⁡π3+12cis⁡2π3+13cis⁡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|+iArgZ)=−log⁡1+itan−1⁡(−3)=itan−1⁡3=iπ3

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

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$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{3}$

answer is D.

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Detailed Solution

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