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 $C = cos \frac{\pi}{3} + \frac{1}{2} cos \frac{2\pi}{3} + \frac{1}{3} cos \frac{3\pi}{3} + \frac{1}{4} cos \frac{4\pi}{3} + ... \infty$
$\therefore C + iS = \left(cos\frac{\pi}{3} + i\,sin\frac{\pi}{3}\right) + \frac{1}{2}\left(cos\frac{2\pi}{3}+ i\,sin\frac{2\pi}{3}\right)$
$ + \frac{1}{3}\left(cos\frac{3\pi}{3} + i\,sin\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 + i\,sin\,\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} - i\,sin\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)$
$ = -log \,z $ (say $z = \frac{1}{2} - i\frac{\sqrt{3}}{2})$
$ = -[log |z| + i\,arg\,z]$,
where $|z| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$
$ = -\left[log\,1 + i\,tan^{1}\left(\frac{Im(z)}{Re\,z}\right)\right]$
$= 0 - i\,tan^{-1}\left(\frac{-\sqrt{3}/2}{1/2} \right)$
$= -i\,tan^{-1}(-\sqrt{3}) = i\,tan^{-1}\left(tan\frac{\pi}{3}\right)$
$\therefore C + iS = i\frac{\pi}{3}$
$\therefore S = \frac{\pi}{3}$