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Question Number 35035 by abdo mathsup 649 cc last updated on 14/May/18

calculate u_n = Σ_(k=1) ^n (k/((k+1)!))

$${calculate}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{k}}{\left({k}+\mathrm{1}\right)!} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18

T_k =(k/((k+1)!)) =((k+1−1)/((k+1)!)) T_k =(1/(k!))−(1/((k+1)!)) T_1 =(1/(1!))−(1/(2!)) T_2 =(1/(2!))−(1/(3!)) T_3 =(1/(3!))−(1/(4!)) ................ ............... T_n =(1/(n!))−(1/((n+1)!)) so T_1 +T_2 +T_3 +....+T_n =(1/(1!))−(1/((n+1)!)) =1−(1/((n+1)!))

$${T}_{{k}} =\frac{{k}}{\left({k}+\mathrm{1}\right)!} \\ $$$$=\frac{{k}+\mathrm{1}−\mathrm{1}}{\left({k}+\mathrm{1}\right)!} \\ $$$${T}_{{k}} =\frac{\mathrm{1}}{{k}!}−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!} \\ $$$${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!} \\ $$$${T}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!} \\ $$$${T}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!} \\ $$$$................ \\ $$$$............... \\ $$$${T}_{{n}} =\frac{\mathrm{1}}{{n}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$${so}\:{T}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} +....+{T}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$

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