Question

The sum to infinite terms
$sin\frac{\pi }{3}+\frac{1}{2}sin\frac{2\pi }{3}+\frac{1}{3}sin\frac{3\pi }{4}+\frac{1}{4}sin\frac{4\pi }{3}+....+\infty$ equals,

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

Answer: (B)

Let $S=sin\frac{\pi }{3}+\frac{1}{2}sin\frac{2\pi }{3}+\frac{1}{3}sin\frac{3\pi }{3}+\frac{1}{4}sin\frac{4\pi }{3}+...\infty$
and $C=cos\frac{\pi }{3}+\frac{1}{2}cos\frac{2\pi }{3}+\frac{1}{3}cos\frac{3\pi }{3}+\frac{1}{4}cos\frac{4\pi }{3}+...\infty$
$\therefore C+iS=\left(cos\frac{\pi }{3}+isin\frac{\pi }{3}\right)+\frac{1}{2}\left(cos\frac{2\pi }{3}+isin\frac{2\pi }{3}\right)$
$+\frac{1}{3}\left(cos\frac{3\pi }{3}+isin\frac{3\pi }{3}\right)+...\infty$
$=e^{i\frac{\pi }{3}}+\frac{1}{2}{e}^{i\frac{2\pi }{3}}+\frac{1}{3}{e}^{i\frac{3\pi }{3}}+...\infty (\because cos\theta +isin\theta =e^{i\theta })$
$=e^{i\frac{\pi }{3}}+\frac{1}{2}(e^{(i\frac{\pi }{3})}{)}^2+\frac{1}{3}(e^{i\frac{\pi }{3}}{)}^3+...\infty$
$=y+\frac{1}{2}{y}^2+\frac{1}{3}{y}^3+...\infty$(say $y=e^{i\frac{\pi }{3}})$
$=-log(1-y)\left(\because \alpha +\frac{{\alpha }^2}{2}+\frac{{\alpha }^3}{3}+...\infty =-log(1-\alpha )\right)$
$=-log(1-e^{i\frac{\pi }{3}})$
$=-log(1-cos\frac{\pi }{3}-isin\frac{\pi }{3})$
$=-log\left[\left(1-\frac{1}{2}\right)-i\frac{\sqrt{3}}{2}\right]=-log\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$
$=-logz$ (say $z=\frac{1}{2}-i\frac{\sqrt{3}}{2})$
$=-[log|z|+iargz]$,
where $|z|=\sqrt{(\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}=1$
$=-\left[log1+ita{n}^1\left(\frac{Im(z)}{Rez}\right)\right]$
$=0-ita{n}^{-1}\left(\frac{-\sqrt{3}/2}{1/2}\right)$
$=-ita{n}^{-1}(-\sqrt{3})=ita{n}^{-1}\left(tan\frac{\pi }{3}\right)$
$\therefore C+iS=i\frac{\pi }{3}$
$\therefore S=\frac{\pi }{3}$