Radius of convergence of a particular power series

Question

What is the radius of convergence of the power series f(x) = $\sum_{n=1}^{\infty}(\log n)x^n$ ?

I want to use root test. So basically how do I find $\lim (\log n)^{1/n}$ ?

Answer 1

Let the radius of convergence be $\rho$, then $\dfrac{1}{\rho}=lim_{k\to\infty} sup\sqrt[n]{log(n)}$

$\forall \varepsilon \gt 0, \exists N\in \Bbb N$ such that $\forall n\ge N, \sqrt[n]{log(n)}\lt L+\varepsilon; \forall n\ge N, \exists m\ge n$ such that $\sqrt[m]{log(m)}\gt L- \varepsilon$

Given any $\varepsilon \in (0,1), \forall n\ge 3 , n\lt 1+n\varepsilon\lt (1+\varepsilon)^n $. Hence, $\sqrt[n]{log(n)}\le \sqrt[n]{n}\lt 1+\varepsilon$. On the other hand, $ 1-\varepsilon\lt 1\lt \sqrt[n]{1}=1.$ Therefore, $\rho =1$.

If $x=1,\sum_{n=1}^{\infty}(\log n)x^n= \sum_{n=1}^{\infty}(\log n)$ diverges since $\sum_{k=1}^{n}log(k)=log(n!)$ is increasing and unbounded. If $x=-1, \sum_{n=1}^{\infty}(\log n)x^n=\sum_{n=1}^{\infty}(\log n)(-1)^n$. As $n\to \infty,(\log n)(-1)^n$ oscillates on $(-\infty, \infty)$. Therefore, the series diverges when $x=-1$ as well.

Answer 2

By L'Hospital,

$$\lim_{n\to\infty}\frac{\log n}{n}=\lim_{n\to\infty}\frac1{n}=0.$$