Question Number 76794 by necxxx last updated on 30/Dec/19
$${Find}\:{the}\:{sum}\:{of} \\ $$$${x}\:+\:\frac{{x}}{\mathrm{1}+{x}}\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:+\:\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+….\:{for} \\ $$$$\mid\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid<\mathrm{1} \\ $$
Commented by turbo msup by abdo last updated on 30/Dec/19
$${S}\left({x}\right)={x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\:+\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \:+….\right) \\ $$$$=\frac{{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}}\:=\frac{{x}}{\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}}=\mathrm{1}+{x} \\ $$
Answered by Rio Michael last updated on 30/Dec/19
$$\:\mathrm{S}_{\infty} \:=\:\frac{{x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}}\:\:=\:\frac{{x}}{\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}}\:=\:{x}\:.\:\frac{\mathrm{1}+{x}}{{x}}\:=\:\mathrm{1}+{x} \\ $$$$ \\ $$