Let S(n)=1+2+3+...+n " and " P(n)=(S(2))...

Let `S_(n)=1+2+3+...+n " and " P_(n)=(S_(2))/(S_(2)-1).(S_(3))/(S_(3)-1).(S_(4))/(S_(4)-1)...(S_(n))/(S_(n)-1)`, where `n inN,(nge2) Then ``underset(ntooo)limP_(n)`=__________.

`1`

`2`

`3`

`4`

Text Solution

Verified by Experts

The correct Answer is:

`(3)`

`S_(n)=(n(n+1))/(n)" and "S_(n)-1=((n+2)(n-1))/(2)`
`:." "(S_(n))/(S_(n)-1)=(n(n+1))/(2).(2)/((n+2)(n-1))=((n)/(n-1))((n+1)/(n+2))`
`:." "P_(n)=((2)/(1).(3)/(2).(4)/(3).(5)/(4)...(n)/(n-1))((3)/(4).(4)/(5).(5)/(6)...(n+1)/(n+2))`
`=((n)/(1))((3)/(n+2))`
`:." "underset(ntooo)limP_(n)=3`

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