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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?

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3 Answers 3

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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!

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  • $\begingroup$ The development of the Fourier series in your answer lacks clarity in my view. For example, the domain over which the Fourier series is valid is not specified nor immediately obvious. I suppose it's $(0,2\pi)$? $\endgroup$ May 2, 2016 at 7:18
  • $\begingroup$ @user5713492 When initially writing, I was viewing $x$ as near $\pi/3$. After reading your comment, I edited to make the development more precise. $\endgroup$
    – Mark Viola
    May 2, 2016 at 15:28
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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.

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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 $$\dfrac\pi3$$

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