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limn→∞0.16log2.5(13+132+133+⋯+13n)= ?\large \lim_{n \to \infty} 0.16^{\log_{2.5}\left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n} \right)} = \, ? n→∞lim0.16log2.5(31+321+331+⋯+3n1)=?
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