How to take limits of 'almost Riemann' sums like lim n → ∞ ∑ k = 0 n 1 n cos ⁡ ( a π k log ⁡ ( n ) / n )

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How to take limits of &#039;almost Riemann&#039; sums like       lim          n      →      ∞            ∑          k      =      0        n        1    n    cos  ⁡  (  a  π  k  log  ⁡  (  n  )      /    n  )

Madelyn Lynch · Accepted Answer

1. Suppose   h  (  n  )      →          n      →      ∞        ∞, and   h  (  n  )  =  o  (  n  )Divide   [  0  ,  h  (  n  )  ] in n subintervals of same size. The size of each subinterval is             h      (      n      )        n   and the Riemann sum of the integral       ∫    0          h      (      n      )        f    d  x for the uniform partition is            h      (      n      )        n        ∑          k      =      1        n    f            (            k    n    h  (  n  )            )      Then      A    n    :=      ∫    0          h      (      n      )        f  (  x  )    d  x  −            h      (      n      )        n        ∑          k      =      1        n    f            (            k    n    h  (  n  )            )        =      ∑          k      =      1              n            ∫                                        k            −            1                    n                    h      (      n      )                                k          n                    h      (      n      )                  (        f  (  x  )  −  f            (                  k      n        h  (  n  )            )                  )          d  x2. Suppose f is uniformly continuous in   [  0  ,  ∞  )Then for any   ε  &amp;gt;  0, there is   N  ∈      N   for which                              1                      h            (            n            )                                    |                          A          n                          |                ≤                  1                      h            (            n            )                                    ∑                      k            =            1                                n                                    ∫                                                                      k                  −                  1                                n                                      h            (            n            )                                                              k                n                                      h            (            n            )                                                |                          f        (        x        )        −        f                              (                                                k            n                          h        (        n        )                              )                                                |                                  d        x        ≤        ε            for   n  ≥  N. That is                    (0)                              1                      h            (            n            )                                    |                          A          n                          |                          →                      n            →            ∞                          0            Notice that                    (1)                              C          n                :=                  1          n                          ∑                      k            =            1                    n                f                              (                                    k          n                h        (        n        )                              )                          =                  1                      h            (            n            )                                                (                                    ∫          0                      h            (            n            )                          f        (        x        )                d        x        −                  A          n                                      )                              3. Let   F  :  x  ↦      ∫    0    x    f  (  t  )    d  t If F is bounded, then from (0) and (1)      C    n        →          n      →      ∞        04. If   f  (  x  )      →          x      →      ∞        b, then from (0) and (1)      C    n        →          n      →      ∞        bConditions 1-3 are satisfied for the function   f  (  x  )  =  cos  ⁡  (  a  π  x  ) and sequence   h  (  n  )  =  log  ⁡  n. Thus,      1    n        ∑          k      =      1        n    cos  ⁡  (  a  π            k      n        log  ⁡  n  )      →          n      →      ∞        0

How to take limits of 'almost Riemann' sums like <munder> <mo movablelimits="true" form="pref

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