Question Number 214310 by malwan last updated on 04/Dec/24
$$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:…\:+\:\frac{\mathrm{99}}{\mathrm{100}!} \\ $$
Answered by mr W last updated on 05/Dec/24
$$\frac{{n}−\mathrm{1}}{{n}!}=\frac{{n}}{{n}!}−\frac{\mathrm{1}}{{n}!}=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\frac{\mathrm{1}}{{n}!} \\ $$$$ \\ $$$${sum}=\left(\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{99}!}−\frac{\mathrm{1}}{\mathrm{100}!}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}!} \\ $$
Commented by malwan last updated on 05/Dec/24
$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$
Answered by Frix last updated on 04/Dec/24
$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{k}}{\left({k}+\mathrm{1}\right)!}\:=\mathrm{1}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$${S}_{\mathrm{99}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}!} \\ $$
Commented by malwan last updated on 05/Dec/24
$${thank}\:{you}\:{sir} \\ $$