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Question Number 33719 by prof Abdo imad last updated on 22/Apr/18

$$simplify{S}_n\left(x\right)=(1+x^2)(1+x^4)....(1+x^{2^n})$$ $$2)find{lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)if\mid x\mid <1.$$

Commented byprof Abdo imad last updated on 23/Apr/18

$$wehave(1+x^2)(1+x^4)=1+x^4+x^2+x^6$$ $$=1+x^2+x^4+x^6=\frac{1-x^8}{1-x^2}if{x}^2\ne 1$$ $$letsupposethat{S}_n\left(x\right)=\frac{1-x^{2^{n+1}}}{1-x^2}$$ $$S_{n+1}\left(x\right)=(1+x^2)(1+x^4).....(1+x^{2^{n+1}})$$ $$=(1+x^{2^{n+1}})S_n\left(x\right)=\frac{(1+x^{2^{n+1}})(1-x^{2^{n+1}})}{1-x^2}$$ $$=\frac{1-{\left(x^{2^{n+1}}\right)}^2}{1-x^2}=\frac{1-x^{2^{n+2}}}{1-x^2}theresultistrueatterm$$ $$n+1andwemuststudytbecasex=\stackrel{-}{+}1$$ $$2)wehaveprovedthat$$ $$S_n\left(x\right)=\frac{1-x^{2^{n+1}}}{1-x^2}sofor\mid x\mid <1{lim}_{n\to +\mathrm{\infty }}{S}_n\left(x\right)=\frac{1}{1-x^2}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18

$$(1+x^2)(1+x^4)=1+x^2+x^4+x^6$$ $$(1+x^2)(1+x^4)(1+x^8)=(1+x^2+x^4+x^6)(1+x^8)$$ $$=1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}$$ $$$$ $$S_n\left(x\right)=(1-x^{2n})/(1-x^2)$$ $$sothevalueof{S}_n\left(x\right)is1/(1-x^2)whenlimitn$$ $$tendstoinfinity$$ $$$$ $$$$ $$$$ $$$$ $$\underset{x\to 0}{\mathrm{i}}$$

Commented byprof Abdo imad last updated on 22/Apr/18

$$itsnotcorrectbecause\frac{1-x^{2n}}{1-x^2}$$ $$=\frac{1-{\left(x^2\right)}^n}{1-x^2}=\frac{(1-x^2)(1+x^2+x^4+...+x^{2n-2})}{1-x^2}$$ $$=1+x^2+x^4+...+x^{2n-2}\ne S_n$$

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