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Question Number 24438 by Tinkutara last updated on 18/Nov/17

$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{to}\mathrm{infinite}\mathrm{terms}\mathrm{of}\mathrm{the}$$$$\mathrm{series}\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+......$$

Commented by prakash jain last updated on 18/Nov/17

$$\mathrm{S}=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+....$$$$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}=\frac{x+x^2+x^3}{1-x^4}$$$$\frac{x+x^2+x^3}{1-x^4}+\frac{x^4}{1-x^8}=\frac{x+x^2+x^3+x^4+x^5+x^6+x^7}{1-x^8}$$$$assuming\mid x\mid <1$$$$\Rightarrow S=\frac{x}{1-x}$$

Commented by Tinkutara last updated on 18/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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