Let L = [n ]
[
] (n! /(mn)n)1/n
Taking log on both sides,
log L = [n ][ ] 1/n[ln( n!/ (mn)n)
=[n ][ ] 1/n [ ln ({ (n/mn) .(n-1/mn).(n-2/mn)...........(2/mn)(1/mn) }]
= [n ][ ] 1/n [ ln(n/mn) +ln( n-1/mn) + .............+ln(2/mn) +ln(1/mn) ]
= [n][ ] [r=0]
[n-1] ln (n-r/mn) (1/n)
=[n][ ] [r=0][n-1] ln(1- (r/n) /m) (1/n)
Now replacing 1/n by integral and r/n by x, we can write the limit as
= [ 0]
[1] ln (1-x/m) dx
Now integration of lnx is xlnx-x, and since the above expression is a linear in x, hence, we can write it as
= -m ( (1-x/m)ln (1-x/m) - (1-x/m)) 01
= ln(1/m) -1
(while applying the limit 1, the form 0*infinity is calculated by using limit, and it comes out to be 0)
Hence L = eln(1/m)-1
=1/me
Moderation Team
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