How can I solve lim n → ∞ 1 n 2 ∑ k = 0 n 1 ln ⁡ ( 1 + ( n + k ) n 2 + k 2 n 3 )

Question

How can I solve      lim          n      →      ∞                  1              n        2                    ∑              k        =        0                    n                            1                  ln          ⁡                      (            1            +                                          (                n                +                k                )                                                      n                    2                                    +                                      k                    2                                                                              n                3                                      )

iberistazi · Accepted Answer

lim          n      →      ∞                  1              n        2                    ∑              k        =        0                    n                            1                  log          ⁡                      (            1            +                                          (                n                +                k                )                                                      n                    2                                    +                                      k                    2                                                                              n                3                                      )                                =      lim          n      →      ∞            1          n      2            ∑          k      =      0        n        1          log      ⁡      (      1      +              1        n            (      1      +              k        n            )              1        +                              k            2                                n            2                              )        .Now, remembering that   x  ≥  log  ⁡  (  1  +  x  ) for every   x  &amp;gt;  0 and that for every   ε  &amp;gt;  0 there is a   δ  &amp;gt;  0 such that for every   x  ∈  (  0  ,  δ  ) we have   log  ⁡  (  1  +  x  )  ≥  (  1  −  ε  )  xThe first inequality implies      lim          n      →      ∞            1          n      2            ∑          k      =      0        n        1          log      ⁡      (      1      +              1        n            (      1      +              k        n            )              1        +                              k            2                                n            2                              )        ≥      lim          n      →      ∞            1          n      2            ∑          k      =      0        n        1                  1        n            (      1      +              k        n            )              1        +                              k            2                                n            2                                =  =      lim          n      →      ∞            1    n        ∑          k      =      0        n        1          (      1      +              k        n            )              1        +                              k            2                                n            2                                =      ∫    0    1        1          (      1      +      x      )              1        +                  x          2                      d  x  .On the other hand, since (for a fixed   ε) for n big enough we have that for every k it holds that       1    n    (  1  +      k    n    )      1    +                  k        2                    n        2              ≤            2              2              n    ≤  δ      lim          n      →      ∞            1          n      2            ∑          k      =      0        n        1          log      ⁡      (      1      +              1        n            (      1      +              k        n            )              1        +                              k            2                                n            2                              )        ≤      1          1      −      ε            ∫    0    1        1          (      1      +      x      )              1        +                  x          2                      d  x  .Since   ε is arbitrary, we get      lim          n      →      ∞            1          n      2            ∑          k      =      0        n        1          log      ⁡      (      1      +              1        n            (      1      +              k        n            )              1        +                              k            2                                n            2                              )        =      ∫    0    1        1          (      1      +      x      )              1        +                  x          2                      d  x  .

How can I solve <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeX

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