Radius and interval of convergence for $\sum_{n=2}^\infty \frac{x^{2n}}{n\ln^2(n)}$ I was requested to show radius and interval of convergence for the series
$$\sum_{n=2}^\infty \frac{x^{2n}}{n\ln^2(n)}$$
and was wondering if my result is correct. This was my approach.
$i)$ $a_n := \frac{x^{2n}}{n\ln^2(n)} = \frac{(x^2)^n}{n\ln^2(n)}$. We have
$$|\frac{a_{n+1}}{a_n}| = \frac{|(x^2)^{n+1}|}{(n+1)\ln^2(n+1)}\times\frac{n \ln^2(n)}{|(x^2)^n|}$$
$$= \frac{n\ln^2(n)}{(n+1) \ln^2(n+1)}|x^2|$$
Because $\lim_{n\to\infty} (\frac{\ln x}{\ln (x+1)})^2 = 1$ (through L'Hopital) we have
$$\lim_{n\to\infty} \frac{n\ln^2(n)}{(n+1) \ln^2(n+1)}= \lim_{n\to\infty} \Big((\frac{\ln(n)}{ln(n+1)})^2\Big) \times \lim_{n\to\infty}\frac{n}{n+1} =1 \times 1=1$$
and therefore $|\frac{a_{n+1}}{a_n}| \to |x^2|$ when $n \to \infty$.
$ii)$ $|x^2|=x^2 <1 \iff x \in(-1, 1) \equiv |x| <1$ . This already implies a radius of convergence of $1$.
$iii)$ $x=1 \implies a_n = \frac{1}{n\ln^2(n)}$. Since
$$\int_{2}^\infty \frac{1}{x \ln^2x} dx= \lim_{t\to\infty}\int_2^t \frac{1}{u^2}du \tag{$u=\ln x$}$$
$$= \lim_{t\to\infty} (-\frac{1}{t} + \frac{1}{2}) = \frac{1}{2}$$
we know the series will converge when $x=1$ via the integral test.
For $x=-1$, $a_n=\frac{((-1)^2)^n}{n \ln^2 n} = \frac{1}{n \ln^2 n}$, the same case we already examined. Therefore,the interval of convergence is $[-1, 1]$.
Final answer.
$R= 1, I=[-1, 1]$.
Is this answer correct?
 A: Your answer is correct, but just for another perspective, note that the $n$'th root test could also be used as:
$$\mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{\frac{{{x^{2n}}}}{{n{{\left( {\ln \left( n \right)} \right)}^2}}}}} = {x^2}\mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{{\frac{1}{{n{{\left( {\ln \left( n \right)} \right)}^2}}}}} = {x^2}$$
so that the region of convergence surely includes:
$${x^2} < 1 \to  - 1 < x <  + 1$$
To check the state of the boundary, your integral test seems satisfactory.
A: In order to see that your series is in fact a power seires about zero, we can do
$$\sum_{n=2}^\infty \frac{x^{2n}}{n\ln^2(n)}=\sum_{n=1}^\infty a_n \,x^n$$
and we thus get:
$$a_n=\begin{cases}0\;,\;\;n=2k-1\;\;\text{is odd}\\{}\\\frac1{k\ln^2k}\;,\;\;n=2k\;\;\text{is even}\end{cases}\;\;,\;\;\;k\in\Bbb N$$
So by Cauchy-Hadamard formula (theorem), with $\;R=\;$ the convergence radius, we get:
$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\sqrt[n]{n\ln^2n}=1\;\implies\;R=1$$
and since we have a Maclaurin series here, the series converges absolutely and uniformly in the interval $\;(-1,1)\;$.
The rest is as you did it (at the end points there is convergence and etc.)
