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Question Number 154328 by liberty last updated on 17/Sep/21

$$S=\frac{1}{1+10}+\frac{2}{1+{10}^2}+\frac{4}{1+{10}^4}+\frac{8}{1+{10}^8}+\dots$$$$$$

Answered by MJS_new last updated on 17/Sep/21

$$\mathrm{having}\mathrm{the}\mathrm{answer}\mathrm{we}\mathrm{could}\mathrm{substract}\mathrm{the}$$$$\mathrm{numbers}\mathrm{of}\mathrm{the}\mathrm{progression}\mathrm{one}\mathrm{by}\mathrm{one}$$$$\mathrm{reaching}\mathrm{zero}\mathrm{not}\mathrm{before}\mathrm{infinity}...$$$$\mathrm{try}\mathrm{with}\frac{1}{9}:$$$$\frac{1}{9}-\frac{1}{11}=\frac{2}{99}$$$$\frac{2}{99}-\frac{2}{101}=\frac{4}{9999}$$$$\frac{4}{9999}-\frac{4}{10001}=\frac{8}{99999999}$$$$...$$$$\frac{2^n}{{10}^{2^n}-1}-\frac{2^n}{{10}^{2^n}+1}=\frac{2^{n+1}}{{10}^{2^{n+1}}-1}>0\mathrm{\forall }n\in \mathbb{N}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{2^{n+1}}{{10}^{2^{n+1}}-1}=0$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}\frac{1}{9}$$

Commented by peter frank last updated on 17/Sep/21

$$\mathrm{more}\mathrm{clarification}\mathrm{please}$$

Commented by liberty last updated on 17/Sep/21

$$why\frac{1}{9}-\frac{1}{2}=\frac{2}{99}?$$$$\frac{1}{9}-\frac{1}{2}=-\frac{7}{18}$$

Commented by MJS_new last updated on 17/Sep/21

$$\mathrm{sorry}\mathrm{typo}$$

Commented by MJS_new last updated on 17/Sep/21

$$\mathrm{inserted}\mathrm{some}\mathrm{lines}.{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{what}\mathrm{I}\mathrm{did}$$

Answered by EDWIN88 last updated on 17/Sep/21

$$Let{S}_1=\frac{1}{1+10}+\frac{2}{1+{10}^2}+\frac{4}{1+{10}^4}+\frac{8}{1+{10}^8}+...$$$$let{S}_2=\frac{1}{1-10}+\frac{2}{1-{10}^2}+\frac{4}{1-{10}^4}+\frac{8}{1-{10}^8}+...$$$$adding{S}_{1}and{S}_{2}gives$$$$\Rightarrow S_1+S_2=\frac{1}{1+10}+\frac{1}{1-10}+\frac{2}{1+{10}^2}+\frac{2}{1-{10}^2}+\frac{4}{1+{10}^4}+\frac{4}{1-{10}^4}+\frac{8}{1+{10}^8}+\frac{8}{1-{10}^8}+...$$$$\Rightarrow S_1+\cancel{S_2}=\cancel{S_2}-\frac{1}{1-10}$$$$\Rightarrow S_1=\frac{1}{10-1}=\frac{1}{9}$$

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