Question Number 24438 by Tinkutara last updated on 18/Nov/17
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinite}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{series}\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }+…… \\ $$
Commented by prakash jain last updated on 18/Nov/17
$$\mathrm{S}=\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }+…. \\ $$$$\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{4}} }=\frac{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$$$\frac{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{8}} }=\frac{{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +{x}^{\mathrm{6}} +{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{8}} } \\ $$$${assuming}\:\mid{x}\mid<\mathrm{1} \\ $$$$\Rightarrow{S}=\frac{{x}}{\mathrm{1}−{x}} \\ $$
Commented by Tinkutara last updated on 18/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$