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Question Number 100304 by Algoritm last updated on 26/Jun/20

Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

t_n =(n/((n+1)!)) = (1/(n!)) −(1/((n+1)!))   and it is a telescopic series

$$\mathrm{t}_{\mathrm{n}} =\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:=\:\frac{\mathrm{1}}{\mathrm{n}!}\:−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\: \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{telescopic}\:\mathrm{series} \\ $$

Commented by Dwaipayan Shikari last updated on 26/Jun/20

T_n =(n/((n+1)!))=(1/(n!))−(1/((n+1)!))  ΣT_n =Σ(1/(n!))−(1/((n+1)!))  S_n =1−(1/(100!))             {General sum=1−(1/((n+1)!))  Suppose you take 1  then it sum is 1−(1/(2!))=(1/2)  {which is (1/(2!))}  taking 2      1−(1/(3!))=(5/6)   {which is (1/(2!))+(2/(3!))=(5/6)}  taking    3      1−(1/(4!))=((23)/(24))  {which is (5/6)+(3/(4!))=((46)/(48))=((23)/(24))

$$\mathrm{T}_{\mathrm{n}} =\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{\mathrm{n}!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!} \\ $$$$\Sigma\mathrm{T}_{\mathrm{n}} =\Sigma\frac{\mathrm{1}}{\mathrm{n}!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!} \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}!}\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{General}\:\mathrm{sum}=\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right. \\ $$$${Suppose}\:{you}\:{take}\:\mathrm{1} \\ $$$${then}\:{it}\:{sum}\:{is}\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}!}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\left\{{which}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}!}\right\} \\ $$$${taking}\:\mathrm{2}\:\:\:\:\:\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}!}=\frac{\mathrm{5}}{\mathrm{6}}\:\:\:\left\{{which}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{3}!}=\frac{\mathrm{5}}{\mathrm{6}}\right\} \\ $$$${taking}\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}!}=\frac{\mathrm{23}}{\mathrm{24}}\:\:\left\{{which}\:{is}\:\frac{\mathrm{5}}{\mathrm{6}}+\frac{\mathrm{3}}{\mathrm{4}!}=\frac{\mathrm{46}}{\mathrm{48}}=\frac{\mathrm{23}}{\mathrm{24}}\right. \\ $$

Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

I think you have to start from n=2. Now do it.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{start}\:\mathrm{from}\:\mathrm{n}=\mathrm{2}.\:\mathrm{Now}\:\mathrm{do}\:\mathrm{it}. \\ $$

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