Find lim n → ∞ 1 n ( n + n − 1 2 + n − 2 3 + ⋯ + 2 n − 1 + 1 n − log ⁡ ( n ! ) )

Question

Find       lim          n      →      ∞            1    n        (    n    +                  n        −        1            2        +                  n        −        2            3        +    ⋯    +          2              n        −        1              +          1      n        −    log    ⁡    (    n    !    )    )

Trey Ross · Accepted Answer

Let       x    n   be the sequence given by                              x          n                                    =                  1          n                          (                      ∑                          k              =              1                        n                                              n              +              1              −              k                        k                    −          log          ⁡          (          k          )          )                                                                =                  1          n                          (                      ∑                          k              =              1                        n                                n            k                    +                      1            k                    −          1          −          log          ⁡          (          k          )          )                                                  (1)                                  =        −        1        +                                                                              1                  n                                                  ∑                                      k                    =                    1                                    n                                                  1                  k                                            ⏟                                            →            0                                    as                                    n            →            ∞                          +                                                                              ∑                                      k                    =                    1                                    n                                                  1                  k                                −                log                ⁡                (                n                )                            ⏟                                            →            γ                                    as                                    n            →            ∞                          −                                                                              1                  n                                                  ∑                                      k                    =                    1                                    n                                log                ⁡                (                k                                  /                                n                )                            ⏟                                            →            −            1                                    as                                    n            →            ∞                              Therefore, we find that      lim          n      →      ∞            x    n    =  γ

Emmy Knox · Accepted Answer

By the Stolz-Cesàro theorem                              lim                      n            →            ∞                                    x          n                                    =                              lim                      n            →            ∞                                    1          n                          ∑                      k            =            1                    n                          (                                    n              +              1              −              k                        k                    −          log          ⁡          (          k          )          )                                                  =                              lim                      n            →            ∞                                                              ∑                              k                =                1                            n                                      (                                                n                  +                  1                  −                  k                                k                            −              log              ⁡              (              k              )              )                                n                                                  =                              lim                      n            →            ∞                                                              ∑                              k                =                1                                            n                +                1                                                    (                                                n                  +                  2                  −                  k                                k                            −              log              ⁡              (              k              )              )                        −                          ∑                              k                =                1                            n                                      (                                                n                  +                  1                  −                  k                                k                            −              log              ⁡              (              k              )              )                                            (            n            +            1            )            −            n                                                            =                              lim                      n            →            ∞                                    ∑                      k            =            1                                n            +            1                                    (                                    n              +              1              −              k                        k                    +                      1            k                    −          log          ⁡          (          k          )          )                −                  ∑                      k            =            1                    n                          (                                    n              +              1              −              k                        k                    −          log          ⁡          (          k          )          )                                                  =                              lim                      n            →            ∞                                    ∑                      k            =            1                                n            +            1                                    1          k                +                  ∑                      k            =            1                                n            +            1                                    (                                    n              +              1              −              k                        k                    −          log          ⁡          (          k          )          )                −                  ∑                      k            =            1                    n                          (                                    n              +              1              −              k                        k                    −          log          ⁡          (          k          )          )                                                  =                              lim                      n            →            ∞                                    ∑                      k            =            1                                n            +            1                                    1          k                −        log        ⁡        (        n        +        1        )                                          =                    γ

Find <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD">

Answered question

Answer & Explanation

New Questions in Calculus 1