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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)`=__________.

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Verified by Experts

The correct Answer is:
3
`S_(n)=(n)/(2)(n+1),S_(n)-1=((n+2)(n-1))/(2)`
`(S_(n))/(S_(n)-1)=(n(n+1))/((n+2)(n-1))=(n)/((n-1))((n+1))/((n+2))`
`P_(n)=((2)/(1),(3)/(2)…(n)/(n-1))((3)/(4).(4)/(5)…(n+1)/(n+2))`
`P_(n)=((n)/(1))((3)/(n+2))impliesunderset(ntoinfty)(lim)P_(n)=3`
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