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Grade: 12th pass
        
please send the detailed solution. In this problem Reiman’s sum concept is used and I don’t know about it.  
one month ago

Answers : (1)

Samyak Jain
326 Points
							
Ans is e^(1/2).
Let the limit be L. Now take natural logarithm on both sides.
lnL  =  lim  (1/n).ln{(1 + 1/n)(1 + 2/n)...(1 + n/n)}
          n\rightarrow\infty
lnL =   lim  (1/n).[ln(1 + 1/n) + ln(1 + 2/n) +...+ ln(1 + n/n)]
          n\rightarrow\infty
 
Now use expansion of ln(1 + x) which is x – x2/2 + x3/3 – … and neglect higher powers of x here as n is tending to infinity.
lnL =   lim  (1/n).[1/n – 1/2n2 + 2/n – 2/2n2 +...+ n/n – 2/n.n2)]
          n\rightarrow\infty
lnL =   lim  (1/n).[(1/n) n(n + 1)/2   –  (1/2n2).n(n + 1)/2]
          n\rightarrow\infty
 
       =  lim  n(n + 1)/2n2   –  lim   n (n + 1)/4n3      =   1/2  –  0   =   1/2
          n\rightarrow\infty                       n\rightarrow\infty
 
\therefore  L  =  e^(1/2)
one month ago
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