George Morales

2022-10-13

Compute Power Series Convergence to a function

Consider the next power series

$$\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}(n){z}^{n}$$

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?

Consider the next power series

$$\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}(n){z}^{n}$$

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

Beginner2022-10-14Added 10 answers

Here is an approach.

Assume $|z|<1$

From $0\le \mathrm{ln}n<n$, $\phantom{\rule{mediummathspace}{0ex}}n=1,2,3,\dots $, we get

$$|\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|\le \sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}|z{|}^{n}\le \sum _{n=1}^{\mathrm{\infty}}n|z{|}^{n}=\frac{|z|}{(1-|z|{)}^{2}}<\mathrm{\infty}$$

Assume $|z|\ge 1$

Then

$$\underset{z\to \mathrm{\infty}}{lim}|\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|=\mathrm{\infty}\ne 0.$$

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

$$\begin{array}{}\text{(1)}& \sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}=\frac{-z\mathrm{log}(1-z)-{z}^{2}\gamma (z)}{1-z}\end{array}$$

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.

Assume $|z|<1$

From $0\le \mathrm{ln}n<n$, $\phantom{\rule{mediummathspace}{0ex}}n=1,2,3,\dots $, we get

$$|\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|\le \sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}|z{|}^{n}\le \sum _{n=1}^{\mathrm{\infty}}n|z{|}^{n}=\frac{|z|}{(1-|z|{)}^{2}}<\mathrm{\infty}$$

Assume $|z|\ge 1$

Then

$$\underset{z\to \mathrm{\infty}}{lim}|\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}|=\mathrm{\infty}\ne 0.$$

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

$$\begin{array}{}\text{(1)}& \sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}n\phantom{\rule{mediummathspace}{0ex}}{z}^{n}=\frac{-z\mathrm{log}(1-z)-{z}^{2}\gamma (z)}{1-z}\end{array}$$

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.

Iris Vaughn

Beginner2022-10-15Added 3 answers

Here is another answer. Clearly, the convergence radius is given by

$$R=\underset{n\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(n)}{\mathrm{ln}(n+1)}=\underset{n\to \mathrm{\infty}}{lim}\frac{n+1}{n}=1$$

Consider now the polylogaritmic function given by

$${\text{Li}}_{s}(z):=\sum _{n=1}^{\mathrm{\infty}}\frac{{z}^{n}}{{n}^{s}}.$$

Then is easily seen that for $z\in \{z\in \mathbb{C}:|z|<1\}$ we have

$$\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}(n){z}^{n}=-{(\frac{\mathrm{\partial}}{\mathrm{\partial}s}{\text{Li}}_{s}(z))}_{s=0}$$

$$R=\underset{n\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(n)}{\mathrm{ln}(n+1)}=\underset{n\to \mathrm{\infty}}{lim}\frac{n+1}{n}=1$$

Consider now the polylogaritmic function given by

$${\text{Li}}_{s}(z):=\sum _{n=1}^{\mathrm{\infty}}\frac{{z}^{n}}{{n}^{s}}.$$

Then is easily seen that for $z\in \{z\in \mathbb{C}:|z|<1\}$ we have

$$\sum _{n=1}^{\mathrm{\infty}}\mathrm{ln}(n){z}^{n}=-{(\frac{\mathrm{\partial}}{\mathrm{\partial}s}{\text{Li}}_{s}(z))}_{s=0}$$

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