It is shown that e 11 12 n n + 1 / 2 e − n &lt; n ! &lt; e n n + 1 / 2 e − n .The question is, how to conclude from here that the limit lim n → ∞ n ! n n + 1 / 2 e − n

Question

It is shown that      e                  11        12                  n          n      +      1              /            2            e          −      n        &amp;lt;  n  !  &amp;lt;  e      n          n      +      1              /            2            e          −      n        .The question is, how to conclude from here that the limit      lim          n      →      ∞                  n      !                      n                  n          +          1                      /                    2                            e                  −          n

Alexia Hart · Accepted Answer

You can prove the monotonicity as follows. Let       a    n    =            n      !                      n                  n          +          1                      /                    2                            e                  −          n                     Then                    a        n                            a                  n          +          1                      =            (              1        +                  1          n                    )              n      +      1              /            2            1    e    =  exp  ⁡      (                  (                  n          +                      1            2                          )            log      ⁡              (                  1          +                      1            n                          )            −      1        )    .But                              (                      n            +                          1              2                                )                log        ⁡                  (                      1            +                          1              n                                )                −        1                            =        (        2        n        +        1        )                  1          2                log        ⁡                  (                                                    1                +                                  1                                      2                    n                    +                    1                                                                              1                −                                  1                                      2                    n                    +                    1                                                                                )                −        1                                          =        (        2        n        +        1        )                  tanh                      −            1                          ⁡                  (                                    1                              2                n                +                1                                              )                −        1        ,            and       tanh          −      1        ⁡  (  x  )  =  x  +                    x        3              3    +                    x        5              5    +  …  &amp;gt;  x for   0  &amp;lt;  x  &amp;lt;  1, i.e.,      (          n      +              1        2              )    log  ⁡      (          1      +              1        n              )    −  1  &amp;gt;  (  2  n  +  1  )      1          2      n      +      1        −  1  =  0.Consequently,                    a        n                            a                  n          +          1                      &amp;gt;      e    0    =  1  ,i.e., the sequence       a    n   is strictly decreasing.

It is shown that e <mrow class="MJX-TeXAtom-ORD"> 11 12 </

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New Questions in Calculus 1