How to take the Limits of Logs log &#x2061;<!-- ⁡ --> ( n

kramberol

kramberol

Answered question

2022-07-14

How to take the Limits of Logs
log ( n ! ) log ( n n )
as n
I believe you have to remove the log raising it to their base. Is this correct ?
Thanks.\

Answer & Explanation

Yair Boyle

Yair Boyle

Beginner2022-07-15Added 10 answers

Hint:
Use Sterling's approximation of n !:
n ! ( n e ) n 2 π n
as n
Solution:
By Sterling's approximation:
lim n log ( n ! ) log ( n n ) = lim n log ( ( n / e ) n 2 π n ) log n n
Use logarithm laws to see
log ( ( n / e ) n 2 π n ) = n log n n + log ( 2 π n )
and log ( n n ) = n log ( n )
Then
n log n n + log ( 2 π n ) n log n = 1 1 log n + log ( 2 π n ) n log n
The last two terms all go to 0 as n . Thus we conclude
lim n log ( n ! ) log ( n n ) = 1
spockmonkey40

spockmonkey40

Beginner2022-07-16Added 4 answers

Hints:
1. ln ( n ! ) = k = 1 n ln ( k ) 1 n ln ( x ) d x = n ln n n + 1.
2. ln ( n n ) = n ln n .

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