Asymptotic behavior of ∑ k = 100 n k α log ⁡ ( k )

Question

Asymptotic behavior of       ∑          k      =      100              n                  k              α                    log      ⁡              (        k        )

Elias Flores · Accepted Answer

Assume that   α  &amp;gt;  −  1. Employing the Euler–Maclaurin formula, making a change of integration variables   t  =  exp  ⁡  (  s      /    (  α  +  1  )  ), and using the known asymptotic expansion of the exponential integral   Ei, we derive                              ∑                      k            =            100                    n                                                                    k                α                                                    log              ⁡              k                                                          =                  ∫                      100                    n                                                                    t                α                                                    log              ⁡              t                                d          t                +                  O                                  (                                                                      n                  α                                                            log                ⁡                n                                              )                =                  ∫                      (            α            +            1            )            log            ⁡            100                                (            α            +            1            )            log            ⁡            n                                                                              e                s                                      s                    d          s                +                  O                                  (                                                                      n                  α                                                            log                ⁡                n                                              )                                                  =        Ei        ⁡        (        (        α        +        1        )        log        ⁡        n        )        −        Ei        ⁡        (        (        α        +        1        )        log        ⁡        100        )        +                  O                                  (                                                                      n                  α                                                            log                ⁡                n                                              )                                                  =        Ei        ⁡        (        (        α        +        1        )        log        ⁡        n        )        +                  O                                  (                                                                      n                  α                                                            log                ⁡                n                                              )                                                  ∼                                            n                              α                +                1                                                          (            α            +            1            )            log            ⁡            n                                    ∑                      m            =            0                    ∞                                                    m              !                                      (              α              +              1                              )                m                                            log                m                            ⁡              n                                      +                  O                                  (                                                                      n                  α                                                            log                ⁡                n                                              )                                                  =                                            n                              α                +                1                                                          (            α            +            1            )            log            ⁡            n                                    [                                    ∑                              m                =                0                            ∞                                                                        m                  !                                                  (                  α                  +                  1                                      )                    m                                                        log                    m                                    ⁡                  n                                                      +                          O                                                  (                                                1                  n                                            )                                ]                                                  ∼                                            n                              α                +                1                                                          (            α            +            1            )            log            ⁡            n                                    ∑                      m            =            0                    ∞                                                    m              !                                      (              α              +              1                              )                m                                            log                m                            ⁡              n                                          as   n  →  +  ∞. Note that the lower limit of the sum does not alter the asymptotic expansion of the sum as long as it it fixed. Also, the asymptotic expansion agrees with that of   Li  ⁡  (      n          α      +      1        )

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