What is the asymptotic behavior of the following series as m → + ∞ ? S m = ∑ r = 0 ∞ r [ ( 1 − 1 2 r + 1 ) m − ( 1 − 1 2 r ) m ] , m ∈ N + .

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What is the asymptotic behavior of the following series as   m  →  +  ∞ ?      S    m    =      ∑          r      =      0        ∞    r      [                  (        1        −                  1                      2                          r              +              1                                      )            m        −                  (        1        −                  1                      2                          r                                      )            m        ]    ,    m  ∈            N        +    .

Myla Pierce · Accepted Answer

Indeed we have      S    m    =      log    2    ⁡  (  m  )  +      O    (  1  )as   m  →  ∞Define       x    r   by      x    r    =  (  1  −      2          −      r            )    m    .Then   r  =      f    m    (      x    r    ) for       f    m   defined by       f    m    (  x  )  =  −      log    2    ⁡  (  1  −      x          1              /            m        ). Now by noting that       f    m   is increasing on   [  0  ,  1  ), we may bound       S    m   by      S    m    =      ∑          r      =      1              ∞            f    m    (      x    r    )  (      x          r      +      1        −      x    r    )  ≤      ∫          0              1            f    m    (  x  )        d    x  .Similarly,      S    m    +  1  =      ∑          r      =      0              ∞        (  r  +  1  )  (      x          r      +      1        −      x    r    )  =      ∑          r      =      0              ∞            f    m    (      x          r      +      1        )  (      x          r      +      1        −      x    r    )  ≥      ∫          0              1            f    m    (  x  )        d    x  .Finally, the integral is computed as                              ∫                      0                                1                                    f          m                (        x        )                          d                x                            =        −                  ∫                      0                                1                          m                  x                      m            −            1                                    log          2                ⁡        (        1        −        x        )                          d                x                                          =                              [            (            1            −                          x              m                        )                          log              2                        ⁡            (            1            −            x            )            ]                                0                                1                          +                  1                      log            ⁡            2                                    ∫                      0                                1                                                1            −                          x              m                                            1            −            x                                            d                x                                          =                              H            m                                log            ⁡            2                          ,            where       H    m    =  1  +      1    2    +  ⋯  +      1    m   is the harmonic number. Therefore, by using the bound   log  ⁡  (  m  )  ≤      H    m    ≤  log  ⁡  (  m  )  +  1, we conclude      log    2    ⁡  (  m  )  −  1  ≤      S    m    ≤      log    2    ⁡  m  +      1          log      ⁡      2        .

What is the asymptotic behavior of the following series as m → + <mi mathv

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