Does the limit exist lim n → ∞ n − H n − ∑ j ≤ n Ω ( j ) j + ∑ j ≤ n ω ( j ) j n ?

Question

Does the limit exist      lim          n      →      ∞                  n      −              H                  n                    −              ∑                  j          ≤          n                                      Ω                      (            j            )                          j            +              ∑                  j          ≤          n                                      ω                      (            j            )                          j              n    ?

Guillemaejy9x · Accepted Answer

Although   Ω  (  j  ) often denotes the number of &#039; factors counted with multiplicity, let&#039;s follow OP&#039;s definition in this answer:  Ω  (  j  )  =      ∑          p              |            j        pThis indicates that      ∑          j      ≤      n                          Ω        (        j        )            j        =      ∑          j      ≤      n            1    j        ∑          p              |            j        p  =      ∑          p      ≤      n        p      ∑                                                      j              ≤              n                                                          p                              |                            j                                                1    j  Let   j  =  p  r, so the rightmost sum becomes   r  ≤  n      /    p                              ∑                      j            ≤            n                                                              Ω              (              j              )                        j                                              =                  ∑                      p            ≤            n                                    ∑                      r            ≤            n                          /                        p                                    1          r                                                  =                  ∑                      p            ≤            n                          log        ⁡                  n          p                +                  O                {        π        (        n        )        }                                          =        π        (        n        )        log        ⁡        n        −        ϑ        (        n        )        +                  O                          (                      n                          log              ⁡              n                                )                    By the &#039; number theorem we know that   ϑ  (  n  )  ∼  π  (  n  )  log  ⁡  n  ∼  n, so we have      ∑          j      ≤      n                          Ω        (        j        )            j        =  o  (  n  )where   π  (  n  ) denotes the number of &#039;s   ≤  n and   ϑ  (  n  )  =      ∑          p      ≤      n        log  ⁡  p. For the second sum, we can use similar tricks to handle the sum over   ω  (  j  )                              ∑                      j            ≤            n                                                              ω              (              j              )                        j                                              =                  ∑                      p            ≤            n                                    1          p                          ∑                      r            ≤            n                          /                        p                                    1          r                                                  =                  ∑                      p            ≤            n                                    1          p                log        ⁡                  n          p                +                  O                          {                      ∑                          p              ≤              n                                            1            p                    }                                                  =                  ∑                      p            ≤            n                                                              log              ⁡              n                        p                          −                  ∑                      p            ≤            n                                                              log              ⁡              p                        p                          +                  O                (        log        ⁡        log        ⁡        n        )                                          =        (        log        ⁡        n        )        log        ⁡        log        ⁡        n        −        log        ⁡        n        +                  O                (        log        ⁡        log        ⁡        n        )        =        o        (        n        )            Since       H    n    =      O    (  log  ⁡  n  )  =  o  (  n  ), we conclude that OP&#039;s limit exists and equals to one.

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New Questions in Calculus 1