"Factorials grow faster than exponential functions, but much slower than double-exponential functions."The author doesn't provide a link let alone a proof of that fact and I don't find it obvious. After all, factorial functions grow really fast but the author says they grow much slower than double exponentials. Does anyone know a proof that lim x → ∞ e e x x ! = ∞

Question

&quot;Factorials grow faster than exponential functions, but much slower than double-exponential functions.&quot;The author doesn&#039;t provide a link let alone a proof of that fact and I don&#039;t find it obvious. After all, factorial functions grow really fast but the author says they grow much slower than double exponentials. Does anyone know a proof that      lim          x      →      ∞                  e                        e                      x                                      x      !        =  ∞

aitorarjolia · Accepted Answer

Simple comparison of factors shows that                    (1)                    n        !        =                  ∏                      k            =            1                    n                k        ≤                  ∏                      k            =            1                    n                n        =                  n          n                    Since                              lim                      n            →            ∞                                    (                      e            n                    −          n          log          ⁡          (          n          )          )                                    =                  lim                      n            →            ∞                                    e          n                          (          1          −                                    n              log              ⁡              (              n              )                                      e              n                                )                                    (2)                                  =        ∞        ⋅        1            we have                              lim                      n            →            ∞                                                e                                          e                n                                                          n            n                                              =                  lim                      n            →            ∞                                    e                                    e              n                        −            n            log            ⁡            (            n            )                                              (3)                                  =        ∞            Applying (1) to (3) gives                    (4)                              lim                      n            →            ∞                                                e                                          e                n                                                          n            !                          =        ∞

Maribel Vang · Accepted Answer

Don&#039;t need Stirling for that. It is much more basic:Let             a              n        +        1                    a              n              =            e                        e                      n                          e                            e                              e            n                              (      n      +      1      )        =      1          n      +      1                  (              e                              e            n                                      )        e                    e                        e          n                      =      1          n      +      1        (      e                  e        n                  )          e      −      1        ≥            e              n        (        e        −        1        )                    n      +      1      . Then            a              n        +        1                    a              n              =            e                        e                      n                          e                            e                              e            n                              (      n      +      1      )        =      1          n      +      1                  (              e                              e            n                                      )        e                    e                        e          n                      =      1          n      +      1        (      e                  e        n                  )          e      −      1        ≥            e              n        (        e        −        1        )                    n      +      1      Now you only need to know that exp grows alot faster than each polynomial. This gives you that even             a              n        +        1                    a      n       grows really fast.

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New Questions in Calculus 1