SolutionSolution by Uettechat · FR0.9999...=0.9+0.09+0.009+...=9(0.1+0.01+0.001+...)=9(10−1+10−2+10−3+...)=9∑n≥110−k=91011−1/10=10.9999...=0.9+0.09+0.009+...=9(0.1+0.01+0.001+...)=9(10^{-1}+10^{-2}+10^{-3}+...)=9\sum_{n\geq1}10^{-k}=\frac{9}{10}\frac{1}{1-1/10}=10.9999...=0.9+0.09+0.009+...=9(0.1+0.01+0.001+...)=9(10−1+10−2+10−3+...)=9n≥1∑10−k=1091−1/101=1
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