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Question Number 214310 by malwan last updated on 04/Dec/24

(1/(2!)) + (2/(3!)) + (3/(4!)) + ... + ((99)/(100!))

$$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:...\:+\:\frac{\mathrm{99}}{\mathrm{100}!} \\ $$

Answered by mr W last updated on 05/Dec/24

((n−1)/(n!))=(n/(n!))−(1/(n!))=(1/((n−1)!))−(1/(n!))    sum=((1/(1!))−(1/(2!)))+((1/(2!))−(1/(3!)))+((1/(3!))−(1/(4!)))+...+((1/(99!))−(1/(100!)))  =1−(1/(100!))

$$\frac{{n}−\mathrm{1}}{{n}!}=\frac{{n}}{{n}!}−\frac{\mathrm{1}}{{n}!}=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}−\frac{\mathrm{1}}{{n}!} \\ $$$$ \\ $$$${sum}=\left(\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+...+\left(\frac{\mathrm{1}}{\mathrm{99}!}−\frac{\mathrm{1}}{\mathrm{100}!}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}!} \\ $$

Commented by malwan last updated on 05/Dec/24

thank you so much sir

$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$

Answered by Frix last updated on 04/Dec/24

S_n =Σ_(k=1) ^n  (k/((k+1)!)) =1−(1/((n+1)!))  S_(99) =1−(1/(100!))

$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{k}}{\left({k}+\mathrm{1}\right)!}\:=\mathrm{1}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$${S}_{\mathrm{99}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}!} \\ $$

Commented by malwan last updated on 05/Dec/24

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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