I'm trying to show lim n → ∞ ( n + n − 1 ) ! ( n ) ! ( n − 1 ) ! 2 n = ∞

Question

I&#039;m trying to show      lim          n      →      ∞                          (                  n                +        n        −        1        )        !                    (                  n                )        !        (        n        −        1        )        !                    2                        n                      =  ∞

Arianna Turner · Accepted Answer

Without Stirling&#039;s approximation, we can do the following: let   p  ≤      n    ≤  p  +  1 Then      lim          n      →      ∞                          (                  n                +        n        −        1        )        !                    (                  n                )        !        (        n        −        1        )        !                    2                        n                      &amp;gt;      lim          n      →      ∞                          (        p        +        n        −        1        )        !                    (        p        +        1        )        !        (        n        −        1        )        !                    2                        n                      =      lim          n      ,      p      →      ∞                  n      ⋅      (      n      +      1      )      ⋅      .      .      .      ⋅      (      n      +      p      −      1      )                      2                  p          +          1                    (      p      +      1      )      !        &amp;gt;      lim          p      →      ∞                          p        2            ⋅      (              p        2            +      1      )      ⋅      .      .      .      ⋅      (              p        2            +      p      −      1      )                      2                  p          +          1                    (      p      +      1      )      !        &amp;gt;      lim          p      →      ∞                  (              p        2            −      1              )                  p                                    2                  p          +          1                    (      p      +      1              )                  p          +          1                      =      lim          p      →      ∞                  (      p      −      1              )                  p                                    2                  p          +          1                    (      p      +      1      )        =  ∞

Landyn Jimenez · Accepted Answer

a    n    =      1          2                        n                                (              n            +      n      −      1      )      !              (              n            )      !      (      n      −      1      )      !      Take logarithms  log  ⁡  (      a    n    )  =  −      n    log  ⁡  (  2  )  +  log  ⁡  (  (      n    +  n  −  1  )  !  )  −  log  ⁡  (  (      n    )  !  )  −  log  ⁡  (  (  n  −  1  )  !  )Use three times Stirling approximation and continue with Taylor series  log  ⁡  (      a    n    )  =      n        (    log    ⁡          (                        n                2            )        +    1    )    +      1    4        (    2    −    log    ⁡          (      4              π        2            n      )        )    +  O      (          1              n              )    log  ⁡  (      a    n    )  =      n        (    log    ⁡          (                        n                2            )        +    1    )    +      1    4        (    2    −    log    ⁡          (      4              π        2            n      )        )    +  O      (          1              n              )  So,   log  ⁡  (      a    n    )  →  ∞

I'm trying to show <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-

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New Questions in Calculus 1