Question Number 86762 by Tony Lin last updated on 31/Mar/20
$${Find}\:{the}\:{sum}\:{of}\:{the}\:{series} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{12}}+\centerdot\centerdot\centerdot \\ $$$${where}\:{the}\:{terms}\:{are}\:{the}\:{reciprocals} \\ $$$${of}\:{the}\:{positive}\:{integers}\:{whose}\:{only}\: \\ $$$${prime}\:{factors}\:{are}\:\mathrm{2}{s}\:{and}\:\mathrm{3}{s} \\ $$
Commented by Prithwish Sen 1 last updated on 31/Mar/20
$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+.......\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+.....\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{x}\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}\:}}\:=\:\mathrm{3}\:\mathrm{please}\:\mathrm{check}. \\ $$
Commented by redmiiuser last updated on 31/Mar/20
$$\mathrm{3} \\ $$$$ \\ $$
Commented by Tony Lin last updated on 31/Mar/20
$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{3} \\ $$
Commented by Prithwish Sen 1 last updated on 31/Mar/20
$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{fix}\:\mathrm{it}\:. \\ $$