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

Find the sum to infinite terms of the series (x/(1−x^2 ))+(x^2 /(1−x^4 ))+(x^4 /(1−x^8 ))+......

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

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

S=(x/(1−x^2 ))+(x^2 /(1−x^4 ))+(x^4 /(1−x^8 ))+.... (x/(1−x^2 ))+(x^2 /(1−x^4 ))=((x+x^2 +x^3 )/(1−x^4 )) ((x+x^2 +x^3 )/(1−x^4 ))+(x^4 /(1−x^8 ))=((x+x^2 +x^3 +x^4 +x^5 +x^6 +x^7 )/(1−x^8 )) assuming ∣x∣<1 ⇒S=(x/(1−x))

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

Commented by Tinkutara last updated on 18/Nov/17

Thank you very much Sir!

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

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