Let ρ ∈ ( 0 , 1 ), and γ ∈ ( 0 , 1 / 2 ). How does one show that 1 t ∑ i = 1 , j = 1 , i ≠ j t ρ | i − j | i γ j γ → 0 ,

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Let   ρ  ∈  (  0  ,  1  ), and   γ  ∈  (  0  ,  1      /    2  ). How does one show that      1    t        ∑          i      =      1      ,      j      =      1      ,      i      ≠      j        t              ρ                        |                i        −        j                  |                                    i        γ                    j        γ              →  0  ,

arhaitategr · Accepted Answer

Grouping terms by   k  :=      |    i  −  j      |  , we see that the sum equals      S          t        (  ρ  )  :=  2      ∑          k      =      1              t      −      1            S          t      ,      k            ρ    k    ,        S          t      ,      k        :=      ∑          j      =      1              t      −      k            1                  j        γ            (      j      +      k              )        γ              .Now, using       ∑          k      =      1        n        k          −      α        ≍      n          1      −      α            /    (  1  −  α  ) for   0  &amp;lt;  α  &amp;lt;  1, we get      lim          t      →      ∞            t          2      γ      −      1            S          t      ,      k        =      1          1      −      2      γ          ⟹        lim          t      →      ∞            t          2      γ      −      1            S    t    (  ρ  )  =      2          1      −      2      γ            ρ          1      −      ρ      by DCT. This is more precise than the needed       S    t    (  ρ  )      /    t  →  0

Let ρ ∈ ( 0 , 1 ) , and γ &#x

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