$\sum_{n=1}^{\infty} \frac{\sin (n x)}{n}$

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∻Solution by Steps∻ ‖ step 1 ⋮ The sum of the series is given by the formula derived from the asksia-ll calculator: $$\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} = \frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}(-1 + e^{ix}))\right)$$ when $ x $ is a real number. ‖ ‖ step 2 ⋮ The formula can be simplified using properties of logarithms: $$\frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}) + \log(-1 + e^{ix})\right)$$ ‖ ‖ step 3 ⋮ Recognize that the complex logarithm can have multiple values due to the periodicity of the exponential function, but the principal value is taken here. ‖ ‖ step 4 ⋮ The approximation of the sum is given by substituting $ e \approx 2.71828 $ and $ i $ as the imaginary unit: $$\approx \frac{1}{2} i \left(\log(1 - 2.71828^{ix}) - \log(2.71828^{-ix}(-1 + 2.71828^{ix}))\right)$$ when $ x $ is a real number. ‖ ∻Answer∻ ⚹ $$\sum_{n=1}^{\infty} \frac{\sin(nx)}{n} = \frac{1}{2} i \left(\log(1 - e^{ix}) - \log(e^{-ix}(-1 + e^{ix}))\right)$$ ⚹ ∻Key Concept∻ ⚹Summation of a trigonometric series involving sine and natural numbers⚹ ∻Explanation∻ ⚹The series is a special case of the Fourier series and can be expressed in terms of complex logarithms. The asksia-ll calculator provides a formula for the sum of the series, which involves complex numbers and the natural logarithm.⚹◊What is the convergence behavior of the series $ \sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$?⍭ Generate me a similar question◊

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