Determine the limit of the sequence ( 1 + 2 k + 3 k + . . . + n k n k + 1 − 1 k + 1 )

Question

Determine the limit of the sequence       (                  1        +                  2          k                +                  3          k                +        .        .        .        +                  n          k                            n                  k          +          1                      −          1              k        +        1              )

extractumzz · Accepted Answer

Note that  n      (                            ∑                      i            =            1                    n                          i          k                            n                  k          +          1                      −          1              k        +        1              )    =  n      (                  (        k        +        1        )                  ∑                      i            =            1                    n                          i          k                −                  n                      k            +            1                                      (        k        +        1        )                  n                      k            +            1                                )    =            (      k      +      1      )              ∑                  i          =          1                n                    i        k            −              n                  k          +          1                            (      k      +      1      )              n        k            Let       a    n    =  (  k  +  1  )      ∑          i      =      1        n        i    k    −  (  k  +  1  )      n          k      +      1       and       b    n    =  (  k  +  1  )      n    k  . Since       b    n   is increasing to   +  ∞, by Stolz Cesaro, we can look at                    a                  n          +          1                    −              a        n                            b                  n          +          1                    −              b        n              =      1          k      +      1                  (      k      +      1      )      (      n      +      1              )                  k                    −      (      n      +      1              )                  k          +          1                    +              n                  k          +          1                            (      n      +      1              )                  k                    −              n        k            Now, by binomial formula, the denominator is of order   k      n          k      −      1        +  o  (      n          k      −      1        ) and the numerator is of order   (  k  +  1  )  k  (  n  +  1      )          k      −      1        −                              (                                      k          +          1                2                              )                      (  n  +  1      )          k      −      1        +  o  (      n          k      −      1        )  =                              (                                      k          +          1                2                              )                      (  n  +  1      )          k      −      1        +  o  (      n          k      −      1        ), hence                    a                  n          +          1                    −              a        n                            b                  n          +          1                    −              b        n              =                                                        (                                                          k              +              1                        2                                              )                                                  (        (        n        +        1                  )                      k            −            1                          +        o        (                  n                      k            −            1                          )        )                    k      (      k      +      1      )              (                  n                      k            −            1                          +        o        (                  n                      k            −            1                          )        )                  →          n      →      ∞                                        (                                      k          +          1                2                              )                          1          k      (      k      +      1      )        =      1    2  and by Stolz-Cesaro             a      n              b      n        →      1    2

Determine the limit of the sequence ( 1 + 2

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