Find the sum of the series: sum from n=1 to infinity (sin
Find the sum of the series: $$ \sum_{n=1}^{\infty} \frac{\sin\frac{3-4n}6\pi}{2n}. $$
Explanation of Method
The argument simplifies via a phase shift to $\frac\pi2-\frac{2n\pi}3$, turning the sine into $\cos\frac{2n\pi}3$. The resulting series $\sum\cos(n\theta)/n$ is a classical Fourier-type sum, evaluated in closed form using the complex-logarithm identity $\sum z^n/n=-\ln(1-z)$ with $z=e^{i\theta}$, then extracting the real part.
- Simplify the sine argument to $\cos\frac{2n\pi}3$ via a phase shift.
- Express $\sum\cos(n\theta)/n$ as $\mathrm{Re}[-\ln(1-e^{i\theta})]$.
- Compute $1-e^{i2\pi/3}$ in polar form and extract the real part.
Step 1 · Simplify the argument
$\dfrac{3-4n}6\pi=\dfrac\pi2-\dfrac{2n\pi}3$, and $\sin\left(\dfrac\pi2-\dfrac{2n\pi}3\right)=\cos\dfrac{2n\pi}3$. So the series equals $\dfrac12\displaystyle\sum_{n=1}^{\infty}\dfrac{\cos\frac{2n\pi}3}n$.
Step 2 · Complex-log representation
With $\omega=e^{i2\pi/3}=-\tfrac12+i\tfrac{\sqrt3}2$: $\displaystyle\sum_{n=1}^{\infty}\frac{\omega^n}n=-\ln(1-\omega)$, so $\displaystyle\sum_{n=1}^{\infty}\frac{\cos\frac{2n\pi}3}n=\mathrm{Re}[-\ln(1-\omega)]$.
Step 3 · Evaluate
$1-\omega=\dfrac32-i\dfrac{\sqrt3}2=\sqrt3\,e^{-i\pi/6}$, so $\ln(1-\omega)=\tfrac12\ln3-i\tfrac\pi6$, giving $-\ln(1-\omega)=-\tfrac12\ln3+i\tfrac\pi6$. Real part: $\sum\cos\frac{2n\pi}3/n=-\tfrac12\ln3$.
Summary Table
| Quantity | Value |
|---|---|
| $1-e^{i2\pi/3}$ (polar) | $\sqrt3\,e^{-i\pi/6}$ |
| $\sum\cos(2n\pi/3)/n$ | $-\tfrac12\ln3$ |
| Final sum | $-\tfrac14\ln3$ |
Additional Notes
- The technique here — converting $\sum\cos(n\theta)/n$ to the real part of $-\ln(1-e^{i\theta})$ — is the same machinery underlying the Fourier series $\sum\sin(nx)/n=\frac{\pi-x}2$ referenced in Problem 133, and extends immediately to any rational-angle trigonometric series with a $1/n$ weight.
- Since $\frac{2\pi}3$ is a period-$3$ angle, the cosine values cycle through $\left(-\frac12,-\frac12,1\right)$, giving an alternative (but more tedious) route via direct grouping by residue class mod $3$, which the complex-exponential method bypasses entirely.
- The appearance of $\ln3$ rather than $\ln2$ (as in the alternating harmonic series of Problems 128, 141) reflects the different angle: period-$2$ oscillation ($\theta=\pi$) gives $\ln2$, while period-$3$ oscillation ($\theta=2\pi/3$) gives $\ln3$ via $|1-e^{i2\pi/3}|=\sqrt3$.
Result
The series sums to: