Radius of convergence of $\sum_{n=1}^{\infty}{\left(\sum_{k=1}^{n}{\frac{1}{k}}\right)}z^n$?

Question

Radius of convergence of $\sum_{n=1}^{\infty}{\left(\sum_{k=1}^{n}{\frac{1}{k}}\right)}z^n$?

This notation really confuses me. Do I have to observe $a_n:=\sum_{k=1}^{n}{\frac{1}{k}}$? And what method should I use then? I usually use Cauchy-Hadamard.

Answer 1

You have correctly interpreted the notation.

You can use the ratio test to find the radius of convergence:

$$\lim\limits_{n\to\infty} \frac{|a_n|}{|a_{n+1}|} = \lim\limits_{n\to\infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\sum_{k=1}^{n+1} \frac{1}{k}} = \lim\limits_{n\to\infty} \frac{\sum_{k=1}^{n+1} \frac{1}{k} - \frac{1}{n+1}}{\sum_{k=1}^{n+1} \frac{1}{k}} = \lim\limits_{n\to\infty} 1 -\frac{1}{(n+1)\sum_{k=1}^{n+1} \frac{1}{k}} = 1$$

Answer 2

The radius of convergence is $1$.

$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{\sum_{k=1}^n\frac1k}=\limsup_{n\to\infty}\sqrt[n]{\ln(n)+\gamma+o(1)}=\limsup_{n\to\infty}\sqrt[n]{\ln(n)},$$ where $\gamma$ is the Euler-Mascheroni constant.

This is equal to $1$ because $$\lim_{n\to\infty}\ln(n)^{1/n}=\lim_{n\to\infty}\exp\left(\frac{\ln(\ln n)}{n}\right)=\exp\left(\lim_{n\to\infty}\frac{\ln(\ln n)}{n}\right)=\exp(0)=1,$$

using l'Hospital in the last step.