Finding radius of convergence of two power series

Since$$\lim_{n\to\infty}\frac{\frac1{(n+1)\log^2(n+1)}}{\frac1{n\log^2n}}=\lim_{n\to\infty}\frac n{n+1}\times\left(\frac{\log n}{\log(n+1)}\right)^2$$and since $\lim_{n\to\infty}\frac n{n+1}=\lim_{n\to\infty}\frac{\log n}{\log(n+1)}=1$, the radius of convergence of your series is equal to $1$. Note that $\log(n+1)=\log(n)+\log\left(\frac{n+1}n\right)$ and that $\lim_{n\to\infty}\log\left(\frac{n+1}n\right)=0$; it's easy to deduce from this that $\lim_{n\to\infty}\frac{\log n}{\log(n+1)}=1$ indeed.

$\lim\limits_{n \to \infty} \frac{n+1}{n} = 1$ and

$$\lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln n} = 1$$

So $$\lim\limits_{n \to \infty} \frac{(n+1)\ln^2(n+1)}{n \ln^2 n} = \left(\lim\limits_{n \to \infty} \frac{n+1}{n}\right) \left(\lim\limits_{n \to \infty} \frac{\ln(n+1)}{\ln n}\right)^2 = 1$$

proving that the convergence radius is equal to $1$.

I would use the $n$th root test:

First series: $$ \sqrt[n]{a_n}=\frac{1}{n^{\ln n/n}}=\exp\left(-\frac{(\ln n)^2}{n}\right)\to \exp(0)=1. $$ Hence, radius of convergence is equal to one.

Second series: $$ \sqrt[n]{a_n}=\frac{1}{\sqrt[n]{n(\ln n)^2}}=\exp\left(-\frac{\ln n+2\ln(\ln n)}{n}\right)\to \exp(0)=1. $$ Again, radius of convergence is equal to one.