Question Number 214471 by issac last updated on 09/Dec/24
$${d}^{\mathrm{2}} −{d}+\mathrm{2}=\mathrm{0} \\ $$$$\underset{{k};\:{d}^{\mathrm{2}} −{d}+\mathrm{2}=\mathrm{0}} {\sum}\:\frac{\mathrm{1}}{{k}}=?? \\ $$
Answered by mr W last updated on 09/Dec/24
$$\mathrm{1}−\frac{\mathrm{1}}{{d}}+\frac{\mathrm{2}}{{d}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Sigma\frac{\mathrm{1}}{{d}}=−\frac{−\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\checkmark \\ $$$$\Pi\frac{\mathrm{1}}{{d}}=\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$