Is this summation solvable? S n = ∑ i = 2 n log i ⁡ ( n ) Is it possible to solve a summation with a variable base of log? S n = ∑ i = 2 n log i ⁡ ( n ) Should I use the derivative of log i ⁡ ( n )

Question

Is this summation solvable?       S    n    =      ∑          i      =      2              n            log    i    ⁡      (    n    )  Is it possible to solve a summation with a variable base of log?      S    n    =      ∑          i      =      2              n            log    i    ⁡      (    n    )  Should I use the derivative of       log    i    ⁡      (    n    )

Collin Gilbert · Accepted Answer

While I don&#039;t believe there is a nice closed form for       S    n  , you can write the sum in terms of known functions and constants up to a very small error. Specifically,      ∑          i      =      2              N            log          i        ⁡  (  N  )  =  li(N)  log  ⁡  N  +  C  log  ⁡  N  +  O  (  1  )  ,where   li  (  N  ) is the logarithmic integral and   C is a constant equal to  C  =      1          log      ⁡      2        +      ∫          2              ∞                  {      x      }              x              log                  2                    ⁡      x        d  x  .Proof: Writing      ∑          i      =      2              N            log          i        ⁡  (  N  )  =  log  ⁡  N      ∑          i      =      2              N            1          log      ⁡      i        ,our goal is then to find an asymptotic for the sum of   1      /    log  ⁡  i. Writing this as a Riemann Stieltjies integral we have      ∑          i      =      2              N            1          log      ⁡      i        =      ∫                  2                  −                                    N                  +                          1          log      ⁡      x        d      [    x    ]    =      ∫          2              N            1          log      ⁡      x        d  x  −      ∫                  2                  −                                    N                  +                          1          log      ⁡      x        d      {    x    }    .By integration by parts,      ∫                  2                  −                                    N                  +                          1          log      ⁡      x        d      {    x    }    =            {      x      }              log      ⁡      x                          |                    x      =              2                  −                            x      =              N                  +                      +      ∫          2              N                  {      x      }              x              log                  2                    ⁡      x        d  x  =      1          log      ⁡      2        +      ∫          2              ∞                  {      x      }              x              log                  2                    ⁡      x        d  x  −      ∫          N              ∞                  {      x      }              x              log                  2                    ⁡      x        d  x  ,and since      ∫          N              ∞                  {      x      }              x              log                  2                    ⁡      x        d  x  =  O      (          1              log        ⁡        N              )    ,we have that      ∑          i      =      2              N            1          log      ⁡      i        =  li  (  N  )  +  C  +  O      (          1              log        ⁡        N              )  where   li  (  N  ) is the logarithmic integral and  C  =      1          log      ⁡      2        +      ∫          2              ∞                  {      x      }              x              log                  2                    ⁡      x        d  x  .Thus it follows that      ∑          i      =      2              N            log          i        ⁡  (  N  )  =  li(N)  log  ⁡  N  +  C  log  ⁡  N  +  O  (  1  )  .(Note that the asymptotic is then       ∑          i      =      2              N            log          i        ⁡  (  N  )  ∼  N  .)Remark: In fact, we could apply integration by parts again to work out the   O  (  1  ) term exactly and evaluate the sum up to an error of   O      (          1      N        )  . This general process of writing the sum       ∑          k      ≤      N        f  (  k  ) as a series whose main term is       ∫          1        N    f  (  x  )  d  x is known as Euler-Maclaurin summation.

deksteenk9 · Accepted Answer

Since       log          i        ⁡  n  =            log      ⁡      n              log      ⁡      i      , we have      S    n    =      ∑          i      =      2              n                  log      ⁡      n              log      ⁡      i        =  log  ⁡  n      ∑          i      =      2              n            1          log      ⁡      i      and we have by Euler-McLaurin summation formula      ∑          i      =      2              n            1          log      ⁡      i        =      ∫          2              n                  d      x              log      ⁡      x        +  log  ⁡      2    n    +  O      (          1              log        ⁡        n              )    ,which leads us to      S    n    =  Li  (  n  )  log  ⁡  n  +      1    2        log    2    ⁡  n  +  log  ⁡      2    log  ⁡  n  +  O  (  1  )  .Hope this helps.

Is this summation solvable? S_n=sum_(i=2)^n log_i(n) Should I use the derivative of log_i(n)?

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