Question

sin pi/3 + (1/2)sin 2pi/3 + (1/3) sin 3pi/3 + ................... infinity

Answer 1

$$\sin \frac {\pi}{3} + \frac {1}{2} \sin \frac {2 \pi}{3} + \frac {1}{3} \sin \frac {3 \pi}{3} + \dots + \frac {1}{n} \sin \frac {n \pi}{3} + \dots$$

**Solution:**

We have

$$\begin{array}{l} \sum_{n=1}^{\infty} \frac{1}{n} \sin \frac{n\pi}{3} = \\ = \sum_{m=0}^{+\infty} \left( \frac{1}{3m+1} \sin \frac{(3m+1)\pi}{3} + \frac{1}{3m+2} \sin \frac{(3m+2)\pi}{3} + \frac{1}{3m+3} \sin \frac{(3m+3)\pi}{3} \right) = \\ = \sum_{m=0}^{+\infty} \left( \frac{1}{3m+1} \sin \left( m\pi + \frac{\pi}{3} \right) + \frac{1}{3m+2} \sin \left( (m+1)\pi - \frac{\pi}{3} \right) + \frac{1}{3m+3} \sin \left( (m+1)\pi \right) \right) = \\ = \sum_{m=0}^{+\infty} \left( \frac{1}{3m+1} (-1)^m \frac{\sqrt{3}}{2} + \frac{1}{3m+2} (-1)^m \frac{\sqrt{3}}{2} + \frac{1}{3m+3} \cdot 0 \right) \\ = \frac{\sqrt{3}}{2} \sum_{m=0}^{+\infty} (-1)^m \left( \frac{1}{3m+1} + \frac{1}{3m+2} \right) \end{array}$$

So we have

$$\sum_{n=1}^{\infty} \frac{1}{n} \sin \frac{n\pi}{3} = \frac{\sqrt{3}}{2} \sum_{m=0}^{+\infty} (-1)^m \left( \frac{1}{3m+1} + \frac{1}{3m+2} \right)$$

Denote

$$a_m = (-1)^m \left( \frac{1}{3m+1} + \frac{1}{3m+2} \right)$$

Because $\frac{1}{3(m+1)+1} < \frac{1}{3m+1}$ and $\frac{1}{3(m+1)+2} < \frac{1}{3m+2}$ then $|a_{m+1}| < |a_m|$.

Because $\lim_{m \to \infty} |a_m| = 0$ then series

$$\frac{\sqrt{3}}{2} \sum_{m=0}^{+\infty} (-1)^m \left( \frac{1}{3m+1} + \frac{1}{3m+2} \right)$$

is convergent. But because harmonic series

$$\sum_{m=0}^{+\infty} \frac{1}{m}$$

is divergent then

$$\sum_{n=1}^{+\infty} \frac{1}{n} \left| \sin \frac{n\pi}{3} \right| \to +\infty$$

Then the series

$$\sin \frac{\pi}{3} + \frac{1}{2} \sin \frac{2\pi}{3} + \frac{1}{3} \sin \frac{3\pi}{3} + \dots + \frac{1}{n} \sin \frac{n\pi}{3} + \dots$$

is conditionally convergent because it is convergent, but not absolutely convergent.