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Question Number 182810 by Mastermind last updated on 14/Dec/22

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{infinite}\mathrm{sum}$$$$(\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{2^3}-\frac{1}{3^3})+...$$$$$$$$$$

Answered by JDamian last updated on 14/Dec/22

$$(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdot \cdot \cdot )-(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdot \cdot \cdot )=$$$$\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2-1}-\frac{1}{3-1}=1-\frac{1}{2}=\frac{1}{2}$$

Answered by HeferH last updated on 14/Dec/22

$$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...=A$$$$\frac{1}{2}(1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}...)=A$$$$1+(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}...)=2A$$$$1+A=2A\Rightarrow A=1$$$$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}...=B$$$$\frac{1}{3}(1+\frac{1}{3}+\frac{1}{3^2}...)=B$$$$1+(\frac{1}{3}+\frac{1}{3^2}...)=3B$$$$1+B=3B\Rightarrow B=\frac{1}{2}$$$$forthequestion:$$$$(\frac{1}{2}-\frac{1}{3})+(\frac{1}{2^2}-\frac{1}{3^2})+(\frac{1}{2^3}-\frac{1}{3^3})+...=C$$$$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}...-(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}...)=C$$$$1-\frac{1}{2}=C$$$$C=\frac{1}{2}$$

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