I have to find the interval of convergence of the following power series: $$\sum_{n=2}^\infty \frac{x^n}{(\ln (n))^2}$$
My approach to the problem:
I start by using the ratio test:
$$\lim_{n \to \infty} \bigg|\frac {a_{n+1}}{a_n}\bigg|$$ $$ = \lim_{n \to \infty} \bigg|\frac{x^{n+1}} {(\ln (n+1))^2} *\frac{(\ln (n))^2}{x^n}\bigg|$$ $$=\lim_{n \to \infty} \bigg(|x| * \bigg(\frac{\ln (n)}{\ln(n+1)}\bigg)^2\bigg)$$ $$= \lim_{n \to \infty}\bigg(|x| *\bigg(\frac {n+1}{n}\bigg)^2\bigg) = |x|$$
By the ratio test, we know that the given series converges when $|x| <1$. However, the ratio test remains inconclusive when $|x| = 1$. We have to consider the cases where $x = \pm 1$.
When $x = 1$, we have that the series become: $\sum_{n=2}^{\infty} {\frac{1}{(\ln (n))^2}} $
When $x = -1$, we have that the series become: $\sum_{n=2}^{\infty} {\frac{(-1)^n}{(\ln (n))^2}}$
I have no idea how to determine the convergence of these two series. I tried the limit comparison test with $1/x$, but end up getting $\infty$, which does not mean anything. Any help on determining such series would be highly appreciated.
We have for all $0\le a<1$
$$\frac {a^n}{\ln^2n}=_\infty o\left(\frac1{n^2}\right)$$ and the Riemann series $\sum\frac1{n^2}$ is convergent so $R\ge1$. Moreover for $a=1$ we have
$$\frac1{\ln^2n}\ge\frac1n,\quad \text{for $n$ large enough}$$ and the harmonic series $\sum \frac1n$ is divergent so $R\le1$ hence $R=1$.