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Question Number 166013 by mathlove last updated on 11/Feb/22

Answered by JDamian last updated on 11/Feb/22

$$x_r=e^{i\frac{\pi }{2^r}}$$$$x_1{x}_2{x}_3\cdot \cdot \cdot x_{\mathrm{\infty }}=e^{i\pi \sum _r\frac{1}{2^r}}=e^{i\pi \cdot 1}=-1$$

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