Calculate below limit lim n → ∞ ( ∑ i = 1 n 1 i − 2 n )

Question

Calculate below limit      lim          n      →      ∞            (          ∑              i        =        1                    n                    1              i              −    2          n        )

Sage Guerra · Accepted Answer

As a consequence of Euler&#039;s Summation Formula, for   s  &amp;gt;  0 we have      ∑          j      =      1        n        1          j      s        =            n              1        −        s                    1      −      s        +  ζ  (  s  )  +  O  (      |        n          −      s            |    )  ,where   ζ is the Riemann zeta function. In your situation,   s  =  1      /    2, so      ∑          j      =      1        n        1          j        =  2      n    +  ζ  (  1      /    2  )  +  O  (      n          −      1              /            2        )  ,and we have the limit      lim          n      →      ∞            (          ∑              j        =        1            n              1              j              −    2          n        )    =      lim          n      →      ∞                  (        ζ  (  1      /    2  )  +  O  (      n          −      1              /            2        )            )        =  ζ  (  1      /    2  )  .

Hailee Stout · Accepted Answer

The following is an elementary consideration, which shows how to compute the limit in terms of an infinite series. It&#039;s evaluation requires usage of Euler&#039;s summation formula, already covered by Dane.Consider the following transformation      ∑          k      =      1        n        1          k        =      ∑          k      =      1        n        (          1              k              −          2                        k                +                  k          +          1                      )    +      ∑          k      =      1        n        2                  k            +              k        +        1            Then use       k    +    1    −      k    =                    (                  k          +          1                −                  k                )                    (                  k          +          1                +                  k                )                            k        +        1            +              k              =            (      k      +      1      )      −      k                      k        +        1            +              k              =      1                  k        +        1            +              k                  ∑          k      =      1        n        1          k        =      ∑          k      =      1        n        1                  k                              (                      k                    +                      k            +            1                    )                2              +  2      ∑          k      =      1        n        (          k      +      1        −          k        )  The latter sum telescopes:      ∑          k      =      1        n        (          k      +      1        −          k        )    =      (          2        −          1        )    +      (          3        −          2        )    +  ⋯  +      (          n      +      1        −          n        )    =      n    +    1    −  1From here:                              (                      ∑                          k              =              1                        n                                1                          k                                )                −        2                  n                                    =                              ∑                      k            =            1                    n                          1                                    k                                                      (                                  k                                +                                  k                  +                  1                                )                            2                                      +        2                  (                      n            +            1                    −                      n                    −          1          )                                                  =                              ∑                      k            =            1                    n                          1                                    k                                                      (                                  k                                +                                  k                  +                  1                                )                            2                                      +        2                  (                      1                                          n                +                1                            +                              n                                              −          1          )                    In the limit:      lim          n      →      ∞            (          ∑              k        =        1            n              1              k              )    −  2      n    =  −  2  +      ∑          k      =      1        ∞        1                  k                              (                      k                    +                      k            +            1                    )                2

Calculate below limit <munder> <mo movablelimits="true" form="prefix">lim <mrow class="M

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