Consider the next power series $$ \sum_{n=1}^{\infty} \ln (n) z^n $$ 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?


Here is an approach.

  • Assume $|z|<1$.

    From $0\leq\ln n<n$, $\:n=1,2,3,\ldots$, we get $$ \left|\sum_{n=1}^{\infty} \ln n\: z^n\right|\leq \sum_{n=1}^{\infty} \ln n\: |z|^n \leq \sum_{n=1}^{\infty} n |z|^n=\frac{|z|}{(1-|z|)^2}<\infty $$

  • Assume $|z|\geq1$.

    Then $$ \lim\limits_{z \to \infty} \left|\ln n \:z^n\right| = \infty \neq 0. $$

This proves that our power series admits a radius of convergence equal to $1$.

Let $z$ be a complex number such that $|z|<1$, then

$$ \sum_{n=1}^{\infty} \ln n\: z^n =\frac{-z\log (1-z)-z^2 \gamma (z)}{1-z} \tag1 $$

where $\gamma(\cdot)$ denotes a special function called the generalized-Euler-constant function which has been studied by Jonathan Sondow and Petros Hadjicostas, amongst others. You will find the relation $(1)$ in this paper (p. $9$).


Here is another answer. Clearly, the convergence radius is given by $$ R=\lim_{n \to \infty} \frac{\ln(n)}{\ln(n+1)} =\lim_{n \to \infty} \frac{n+1}{n}= 1 $$ Consider now the polylogaritmic function given by $$ \text{Li}_s(z):=\sum_{n=1}^{\infty} \frac{z^n}{n^s}. $$ Then is easily seen that for $z \in \left\{ z \in \mathbb{C} : |z|<1 \right\}$ we have $$ \sum_{n=1}^{\infty} \ln(n) z^n = - \left(\frac{\partial}{\partial s} \text{Li}_s(z) \right)_{s=0} $$


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