Question Number 81768 by Power last updated on 15/Feb/20
Commented by mathmax by abdo last updated on 15/Feb/20
$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}}{\left({k}+\mathrm{1}\right)!}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{k}+\mathrm{1}−\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:=\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{1}}{{k}!}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}!}\:+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\:+...+\frac{\mathrm{1}}{{n}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}\:=\mathrm{1}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$
Commented by mr W last updated on 15/Feb/20
$${answer}\:{see}\:{Q}\mathrm{81506}! \\ $$
Commented by Power last updated on 15/Feb/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by abdomathmax last updated on 15/Feb/20
$${you}\:{are}\:{welcome} \\ $$
Answered by mr W last updated on 15/Feb/20
$$=\mathrm{1}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$
Commented by mr W last updated on 15/Feb/20
$${sir}:\:{in}\:{Q}\mathrm{81506}\:{i}\:{gave}\:{you}\:{all}\:{the}\:{details} \\ $$$${for}\:{the}\:{proof}. \\ $$
Commented by Power last updated on 15/Feb/20
$$\mathrm{sir}\:\mathrm{prove}\:\mathrm{thst}\:\mathrm{pls} \\ $$