Radius and interval of convergence?

Question

I can't figure out how to attack this problem. Do I use the ratio test?

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

Answer 1

Obeserve there's a way to make your series an actual power series:

$$a_n:=\begin{cases}\cfrac1{k\log^2k},&\exists\,1<k\in\Bbb N\;\;s.t.\;\;n=2^k\\{}\\0,&\text{otherwise}\end{cases}$$

and then we get that

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

and we can get the radius of convergence with the Cauchy-Hadamard Theorem (Formula):

$$\frac1R=\overline{\lim_{n\to\infty}}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\frac1{\sqrt[n]{n\log^2n}}=1\implies R=1$$

and the convergence interval is $\;[-1,1]\;$

Answer 2

Hint. Use the root test. Evaluate the limit $$\lim_{n\to\infty}\left(\frac{1}{n\ln^2(n)}\right)^{1/2^n}= \lim_{n\to\infty}\exp\left(-\frac{\ln(n\ln^2(n))}{2^n}\right).$$