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.

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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.

Yair Boyle · Accepted Answer

Hint:Use Sterling&#039;s approximation of   n  !:  n  !  ∼            (              n        e            )        n        2    π    n  as   n  →  ∞Solution:By Sterling&#039;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 · Accepted Answer

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  .

How to take the Limits of Logs log ⁡ ( n

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