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cacul-x-0-1-n-0-nx-n-




Question Number 181207 by lapache last updated on 23/Nov/22
cacul  ∀x∈]0,1[  Σ_(n=0) ^(+∞) nx^n
$${cacul} \\ $$$$\left.\forall{x}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{nx}^{{n}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22
S=x+2x^2 +3x^3 +4x^4 +∙∙∙  xS=x^2 +2x^3 +3x^4 +∙∙∙  S−xS=x+x^2 +x^3 +x^4 +∙∙∙=(x/(1−x))  (1−x)S=(x/(1−x)) ⇒S=(x/((1−x)^2 ))
$${S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$${xS}={x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$${S}−{xS}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot=\frac{{x}}{\mathrm{1}−{x}} \\ $$$$\left(\mathrm{1}−{x}\right){S}=\frac{{x}}{\mathrm{1}−{x}}\:\Rightarrow{S}=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$

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