How do I evaluate this limit? lim n → + ∞ ∑ k = 1 n 1 k ( k + 1 ) ⋯ ( k + m ) ( m = 1 , 2 , 3 , ⋯ )

Question

How do I evaluate this limit?      lim          n      →      +      ∞            ∑          k      =      1              n            1          k      (      k      +      1      )      ⋯      (      k      +      m      )          (  m  =  1  ,  2  ,  3  ,  ⋯  )

Aiden Norman · Accepted Answer

Rewriting the term in the sum with factorials, notice that      1          k      (      k      +      1      )      ⋯      (      k      +      m      )        =      1          m      !            (                  (        k        −        1        )        !        m        !                    (        k        +        m        )        !              )    .Then since            (      k      −      1      )      !      m      !              (      k      +      m      )      !        =      ∫    0    1        x          k      −      1        (  1  −  x      )    m    d  x  ,which can be proved by induction or by using a property of the Beta Function, we see that      ∑          k      =      1        ∞        1          k      (      k      +      1      )      ⋯      (      k      +      m      )        =      ∑          k      =      1        ∞        (          1              m        !                    ∫      0      1              x              k        −        1              (    1    −    x          )      m        d    x    )    =      1          m      !            ∫    0    1    (  1  −  x      )    m        (          ∑              k        =        1            ∞              x              k        −        1              )    d  x  =      1          m      !            ∫    0    1    (  1  −  x      )          m      −      1        d  x  =      1          m      !      m        .

dourtuntellorvl · Accepted Answer

Put       s          n      ,      m        =      ∑          k      =      1        n        1          k      (      k      +      1      )      …      (      k      +      m      )      . We have for   m  ≥  2                              s                      n            ,            m                                              =                  1          m                          ∑                      k            =            1                    n                                      k            +            m            −            k                                k            (            k            +            1            )            …            (            k            +            m            )                                                            =                  1          m                          (                      s                          n              ,              m              −              1                                −                      ∑                          j              =              2                                      n              +              1                                            1                          j              …              (              j              −              1              +              m              )                                )                                                  =                  1          m                          (                      s                          n              ,              m              −              1                                −                      s                          n              +              1              ,              m              −              1                                +                      1                          1              ⋯              (              1              −              1              +              m              )                                )                                                  =                                            s                              n                ,                m                +                1                                      −                          s                              n                +                1                ,                m                −                1                                      +                          1                              m                !                                              m                ,            and since the series       ∑          k      ≥      1            1          k      (      k      +      1      )      …      (      k      +      m      −      1       is convergent,       lim          n      →      ∞            s          n      ,      m      +      1        −      s          n      +      1      ,      m      −      1        =  0 so       lim          n      →      ∞            s          n      ,      m        =      1          m      ⋅      m      !      . This formula also works for   m  =  1 by a direct computation.

How do I evaluate this limit? <munder> <mo movablelimits="true" form="prefix">lim <mrow

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