What interval does this series converge for?

Question

$$\sum_{n=2}^\infty \frac{x^n}{\frac{n}{\ln n}^\frac{n}{\ln n}}$$

What values of $x$ does this series converge for? And is there another formula for this series that can be used to make an analytic continuation of it?

Answer 1

We have to find the radius of convergence of the given power series. Compairing the given series with $\sum_{n=2}^{\infty}a_nx^n$, we get $\sqrt[n]{a_n}=\Big(\frac{\ln n}{n}\Big)^{\frac{1}{\ln n}}$, i.e. $\ln (\sqrt[n]{a_n})=\frac{\ln(\ln n)}{\ln n}-1$, and applying L'Hospital we see that this converges to $-1$. Hence $\sqrt[n]{a_n}\rightarrow \frac{1}{e}$ as $n\rightarrow\infty$. So the radius of convergence is $R=\frac{1}{\limsup \sqrt[n]{a_n}}=e$, and the series converges in $\{z\in\mathbb{C}:|z|<e\}$.