Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$
I have to find the imaginary part of $$P=\sum_{r=1}^\infty \frac{1}{r}z^r$$
Let $$S=1+z+z^2+\cdots$$
Hence, $$P=\int_0^z Sdz=z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots$$
which is the required sum.
$$S=\frac{1}{1-z}$$
$$P=\int_0^z \frac{1}{1-z}dz$$
$$P=\ln\left(\frac{1}{1-z}\right)=\ln\left(\frac{1}{1-e^{i\frac{\pi}{3}}}\right)=i\frac{\pi}{3}$$
Hence, the imaginary part of $P$ is $\frac{\pi}{3}$
Is this method correct? Is there any method that does not require complex numbers?
 A: This can be done with Fourier series. Let $f(x)=x$ for $x\in(-\pi,\pi)$, and its periodic extension with period $2\pi$. Then, since $f(x)=-f(-x)$ we can represent it as
$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(nx\right)$$
Then we can compute
$$\begin{align}\int_{-\pi}^{\pi}f(x)\sin(nx)dx&=\int_{-\pi}^{\pi}x\sin(nx)dx=\left[-\frac xn\cos(nx)+\frac1{n^2}\sin(nx)\right]_{-\pi}^{\pi}\\
&=-\frac{2\pi}n(-1)^n=\int_{-\pi}^{\pi}\sum_{m=1}^{\infty}b_m\sin(mx)\sin(nx)dx\\
&=\sum_{m=1}^{\infty}b_m\pi\delta_{mn}=\pi b_n\end{align}$$
So
$$f(x)=\sum_{n=1}^{\infty}\frac2n(-1)^{n+1}\sin(nx)$$
We can see that
$$\sin\left(n\left(\pi-\frac{\pi}3\right)\right)=\sin(n\pi)\cos\left(n\frac{\pi}3\right)-\cos(n\pi)\sin\left(n\frac{\pi}3\right)=(-1)^{n+1}\sin\left(n\frac{\pi}3\right)$$
And finally
$$\frac{2\pi}3=\pi-\frac{\pi}3=f\left(\pi-\frac{\pi}3\right)=\sum_{n=1}^{\infty}\frac2n(-1)^{n+1}\sin\left(n\left(\pi-\frac{\pi}3\right)\right)=\sum_{n=1}^{\infty}\frac2n\sin\left(n\frac{\pi}3\right)$$
Which agrees with the result in the question.
A: The analysis in the OP works.  Here, we address the question regarding the necessity of using complex analysis.  
To that end, we first note that using standard trigonometric identities, we can evaluate the sum
$$\begin{align}\sum_{n=1}^N \cos(nx)&=\frac{\sin(x)}{\sin(x)}\sum_{n=1}^N \cos(nx)\\\\
&=\csc(x)\sum_{n=1}^N\left(\frac{\sin((n+1)x)-\sin((n-1)x)}{2}\right)\\\\
&=\csc(x)\left(\frac{\sin((N+1)x)+\sin(Nx)-\sin(x)}{2}\right)\\\\
&=\csc(x)\left(\sin\left(\frac{(2N+1)x}{2}\right)\cos\left(\frac{x}{2}\right)\right)-\frac12\\\\
&=\frac{\sin\left(\frac{(2N+1)x}{2}\right)}{2\sin(x/2)}-\frac12
\end{align}$$
Then, we can write the series of interest as
$$\begin{align}
\sum_{n=1}^\infty \frac{\sin(nx)}{n}&=\lim_{N\to \infty}\int_0^x \sum_{n=1}^N \cos(nx') \,dx'\\\\
&=\lim_{N\to \infty}\int_0^x \frac{\sin\left(\frac{(2N+1)x'}{2}\right)}{2\sin(x'/2)} \,dx'-\frac12 x \tag 1\\\\
\end{align}$$
Finally, evaluating the limit in $(1)$ reveals for $|x|<2\pi$
$$\begin{align}
\lim_{N\to \infty}\int_0^x \frac{\sin\left(\frac{(2N+1)x'}{2}\right)}{2\sin(x'/2)} \,dx'&=\lim_{N\to \infty}\int_0^{(N+1/2)x} \frac{\sin(x')}{(2N+1)\sin\left(\frac{x'}{2N+1}\right)} \,dx' \\\\
&=\text{sgn}(x)\int_0^\infty \frac{\sin(x)}{x}\,dx \\\\
&=\frac{\pi}{2}\text{sgn}(x)\tag 2
\end{align}$$
Putting $(1)$ and $(2)$ together yields for $0<x<2\pi$
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac12(\pi-x)$$
Therefore, for $x=\pi/3$, we find that
$$\sum_{n=1}^\infty \frac{\sin(n\pi/3)}{n}=\frac{\pi}{3}$$
as expected!
A: In your line 
$$P=-\ln(1-e^{i\pi/3})$$
Now $$\ln(1-e^{i\pi/3})=\ln(-1)+\ln(e^{i\pi/6})+\ln(2i\sin\dfrac\pi6)\equiv i\pi+\dfrac{i\pi}6+\ln(i)\pmod{2\pi i}$$
$$\equiv i\left(\pi+\dfrac\pi6+\dfrac\pi2\right)\equiv-\dfrac{i\pi}3$$
So, the principal value of the imaginary part of $P$ is $$\dfrac\pi3$$
