How to evaluate : lim n → ∞ ∫ 0 n ( 1 − x n ) n e x 2 d x

Question

How to evaluate :      lim          n      →      ∞            ∫    0    n              (      1      −              x        n            )        n        e                  x        2              d  x

exalantaswo · Accepted Answer

Use the fact that            (      1      −                        x          n                    )        n    ≤      e          −      x      i.e.            (      1      −                        x          n                    )        n        e          x              /            2        ≤      e          −      x              /            2      and      lim          n      →      ∞                  (      1      −                        x          n                    )        n    =      e          −      x      Consider the sequence      f    n    (  x  )  =      {                                                      (              1              −                                                x                  n                                            )                        n                                e                          x                              /                            2                                                x          ∈          [          0          ,          n          ]                                      0                          x          &amp;gt;          n                        which is dominated by   g  (  x  )  =      e          −      x              /            2      . Now apply dominated convergence theorem to get that      lim          n      →      ∞            ∫    0    n        f    n    (  x  )  d  x  =      lim          n      →      ∞            ∫    0          ∞            f    n    (  x  )  d  x  =      ∫    0          ∞            lim          n      →      ∞            f    n    (  x  )  d  x  =      ∫    0          ∞            e          −      x              /            2        d  x  =  2

Christopher Saunders · Accepted Answer

For   x  =  o  (  n  ) and n large enough we have  n  log  ⁡      (    1    −          x      n        )    =  −      ∑          k      =      1              ∞                  x      k              k              n                  k          −          1                      ≥  −  x  −            x      2        n    ,so that            (      1      −              x        n            )        n    ≥      e          −      x      −              x        2                    /            n        .This gives                              ∫          0          n                                      (            1            −                          x              n                        )                    n                          e                      x                          /                        2                                  d        x                            ≥                  ∫          0                                    n                              1                                  /                                4                                                                          (            1            −                          x              n                        )                    n                          e                      x                          /                        2                                  d        x                                          ≥                  ∫          0                                    n                              1                                  /                                4                                                              e                      −            x                          /                        2            −                          x              2                                      /                        n                                  d        x                                          ≥                  e                      −            1                          /                                      n                                                ∫          0                                    n                              1                                  /                                4                                                              e                      −            x                          /                        2                                  d        x        .            The last expression converges to      ∫    0    ∞        e          −      x              /            2          d  x  =  2as   n  →  ∞. Marvis&#039; answer shows that      ∫    0    n              (      1      −              x        n            )        n        e          x              /            2          d  x  ≤      ∫    0    ∞        e          −      x              /            2          d  x  =  2  ,so we conclude from the squeeze theorem that      ∫    0    n              (      1      −              x        n            )        n        e          x              /            2          d  x  ⟶  2as   n  →  ∞.

How to evaluate : lim_(n-> inffty) (1- x/n)^n e^(x/2)dx

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