Compute Power Series Convergence to a functionConsider the next power series ∑ n = 1 ∞ ln ⁡ ( n ) z n Find the convergence radius and a the function f to which the series converges.I have easily found that R = 1 is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?

Question

Compute Power Series Convergence to a functionConsider the next power series      ∑          n      =      1              ∞        ln  ⁡  (  n  )      z    n  Find the convergence radius and a the function  f to which the series converges.I have easily found that   R  =  1 is the convergence radius, however I can not find the function. I was trying to found an elemental function with this power series expantion, but I have failed. Anyone knows such function and how to prove the convergence?

blogpolisft · Accepted Answer

Here is an approach.Assume       |    z      |    &amp;lt;  1From   0  ≤  ln  ⁡  n  &amp;lt;  n,     n  =  1  ,  2  ,  3  ,  …, we get      |          ∑              n        =        1                    ∞              ln    ⁡    n              z      n        |    ≤      ∑          n      =      1              ∞        ln  ⁡  n        |    z            |        n    ≤      ∑          n      =      1              ∞        n      |    z            |        n    =                    |            z              |                    (      1      −              |            z              |                    )        2              &amp;lt;  ∞Assume       |    z      |    ≥  1Then      lim          z      →      ∞            |    ln    ⁡    n              z      n        |    =  ∞  ≠  0.This proves that our power series admits a radius of convergence equal to 1.Let   z be a complex number such that       |    z      |    &amp;lt;  1, then                    (1)                              ∑                      n            =            1                                ∞                          ln        ⁡        n                          z          n                =                              −            z            log            ⁡            (            1            −            z            )            −                          z              2                        γ            (            z            )                                1            −            z                              where   γ  (  ⋅  ) denotes a special function called the generalized-Euler-constant function which has been studied by Jonathan Sondow and Petros Hadjicostas, amongst others.

Iris Vaughn · Accepted Answer

Here is another answer. Clearly, the convergence radius is given by  R  =      lim          n      →      ∞                  ln      ⁡      (      n      )              ln      ⁡      (      n      +      1      )        =      lim          n      →      ∞                  n      +      1        n    =  1Consider now the polylogaritmic function given by      Li    s    (  z  )  :=      ∑          n      =      1              ∞                  z      n              n      s        .Then is easily seen that for   z  ∈      {    z    ∈          C        :          |        z          |        &amp;lt;    1    }   we have      ∑          n      =      1              ∞        ln  ⁡  (  n  )      z    n    =  −            (              ∂                  ∂          s                            Li        s            (      z      )      )              s      =      0

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