How can I get an approximation formula for the sum J ( n ) = 2 − n ∑ k = 1 n 1 k ( n k ) ?

Question

How can I get an approximation formula for the sum   J  (  n  )  =      2          −      n            ∑          k      =      1        n        1    k                                (                            n        k                              )                    ?

Oliver Shepherd · Accepted Answer

(1)                              2                      −            n                                    ∑                      k            =            1                    n                          1          k                                                    (                                            n            k                                              )                                                          =                  2                      −            n                                    ∑                      k            =            1                    n                          1          k                          ∑                      j            =            k                    n                                                    (                                                          j              −              1                                      k              −              1                                                          )                                                          (2)                                  =                  2                      −            n                                    ∑                      k            =            1                    n                          ∑                      j            =            k                    n                          1          j                                                    (                                            j            k                                              )                                                          (3)                                  =                  2                      −            n                                    ∑                      j            =            1                    n                          ∑                      k            =            1                    j                          1          j                                                    (                                            j            k                                              )                                                          (4)                                  =                  2                      −            n                                    ∑                      j            =            1                    n                                                    2              j                        −            1                    j                                    (5)                                  ∼                  ∑                      j            =            0                                n            −            1                                                2                          −              j                                            n            −            j                                              (6)                                  =                  1          n                          ∑                      j            =            0                                n            −            1                                    2                      −            j                                    ∑                      k            =            0                    ∞                                      j            k                                n            k                                              (7)                                  ∼                  ∑                      k            =            0                    ∞                          1                      n                          k              +              1                                                ∑                      j            =            0                    ∞                                      j            k                                2            j                                              (8)                                  =                  ∑                      k            =            0                    ∞                          1                      n                          k              +              1                                                ∑                      j            =            0                    ∞                          ∑                      i            =            0                    k                          2                      −            j                                                              (                                            j            i                                              )                                                {                      k            i                    }                i        !                            (9)                                  =                  ∑                      k            =            0                    ∞                          1                      n                          k              +              1                                                ∑                      i            =            0                    k                          {                      k            i                    }                i        !                  ∑                      j            =            0                    ∞                          2                      −            j                                                              (                                            j            i                                              )                                                          (10)                                  =                  ∑                      k            =            0                    ∞                          2                      n                          k              +              1                                                ∑                      i            =            0                    k                          {                      k            i                    }                i        !                            (11)                                  =                  ∑                      k            =            0                    ∞                                      2                          a              k                                            n                          k              +              1                                                          (12)                                  =                  2          n                +                  2                      n            2                          +                  6                      n            3                          +                  26                      n            4                          +                  150                      n            5                          +        O                          (                      1                          n              6                                )                                    (13)                                  =                  2                      n            −            1                          +                  4                      (            n            −            1                          )              3                                      +                  12                      (            n            −            1                          )              4                                      +                  76                      (            n            −            1                          )              5                                      +        O                          (                      1                          n              6                                )

How can I get an approximation formula for the sum J ( n ) = 2 <mrow c

Answered question

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