Logarithms, prove this limit.Mathematica knows that: log ⁡ ( n ) = lim s → 1 ( 1 − 1 n s − 1 ) ζ ( s )Kind of tautological starting with logarithms, but I would like to know better why this limit works: log ⁡ ( n ) = lim s → 1 + 2 i π k log ⁡ ( n ) ( 1 − 1 n s − 1 ) ζ ( s − i ℑ ( s ) ) for k an integer.Solving it symbolically for some integer k while leaving n as a variable Mathematica says it is equal to zero. But setting n to any value I get log ⁡ ( n )

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Logarithms, prove this limit.Mathematica knows that:  log  ⁡  (  n  )  =      lim          s      →      1              (    1    −          1              n                  s          −          1                      )    ζ  (  s  )Kind of tautological starting with logarithms, but I would like to know better why this limit works:            log      ⁡      (      n      )      =              lim                  s          →          1          +                                    2              i              π              k                                      log              ⁡              (              n              )                                                        (        1        −                  1                      n                          s              −              1                                      )            ζ      (      s      −      i      ℑ      (      s      )      )      for   k an integer.Solving it symbolically for some integer   k while leaving   n as a variable Mathematica says it is equal to zero. But setting   n to any value I get   log  ⁡  (  n  )

humbast2 · Accepted Answer

I have no idea what Mathematica is up to, but if we write   s  =  σ  +            2      i      π      k              log      ⁡      n        +  i  t, we have                              1                      n                          s              −              1                                                          =        exp        ⁡                  (          (          1          −          s          )          log          ⁡          n          )                                                  =        exp        ⁡                  (                      (            1            −            σ            −                                          2                i                π                k                                            log                ⁡                n                                      −            i            t            )                    log          ⁡          n          )                                                  =        exp        ⁡                  (          (          1          −          σ          −          i          t          )          log          ⁡          n          )                ⋅                  e                      −            2            i            π            k                                                            =                  e                      (            1            −            σ            −            i            t            )            log            ⁡            n                                                            =        1        −        (        σ        −        1        +        i        t        )        log        ⁡        n        +        O                  (          (          σ          −          1          +          i          t                      )            2                    )                .            With  ζ  (  s  −  i  ℑ  s  )  =  ζ  (  σ  )  =      1          σ      −      1        +  O  (  1  )  ,we therefore have                              (          1          −                      1                          n                              s                −                1                                              )                ζ        (        s        −        i        ℑ        s        )                            =                  (          (          σ          −          1          +          i          t          )          log          ⁡          n          +          O                      (            (            σ            −            1            +            i            t                          )              2                        )                    )                          (                      1                          σ              −              1                                +          O          (          1          )          )                                                  =        log        ⁡        n        +                              i            t            log            ⁡            n                                σ            −            1                          +        O                  (                                    t              2                                      σ              −              1                                )                +        o        (        1        )        ,            and the limit does not generally exist. The limit exists and equals   log  ⁡  n if   s approaches   1  +            2      i      π      k              log      ⁡      n       in such a way that  lim      t          σ      −      1        =  0  ,but along the ray   t  =  c  (  σ  −  1  ), the limit is   (  1  +  i  c  )  log  ⁡  n, and if   t approaches   0 slower than   σ  −  1, the expression is unbounded.Apparently, for the nonexisiting limit, Mathematica just returned   0 in the general case, but what made it produce the limit   log  ⁡  n when given a specific   n, I cannot guess.

Logarithms, prove this limit. Mathematica knows that: log ⁡ ( n ) =

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