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

Commented byPRITHWISH SEN 2 last updated on 26/Jun/20

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

Commented byDwaipayan 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))

Commented byPRITHWISH SEN 2 last updated on 26/Jun/20

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