Finding radius of convergence of two power series I have two series (1) $\sum \frac{x^n}{n^{\log n}}$ and (2) $\sum \frac{x^n}{n (\log n)^2}$. I need to find the radius of convergence for both the cases.
Attemp: (1) i know i have to find this limit i.e
$$\lim_{n\to \infty}\left |\frac{a_{n+1}}{a_n} \right|=\lim_{n\to \infty}\frac{n^{\log n}}{(n+1)^{\log (n+1)}}=\lim_{n\to \infty}e^{\log^2 n-\log^2 (n+1)}\\=\lim_{n\to \infty} \left(1+ \mathcal{O}\left(\frac{1}{n}\right)\right) \to 1$$ So $R=1$ for this case.
(2) $$L=\lim_{n\to \infty}\left |\frac{a_{n+1}}{a_n} \right|=\lim_{n\to \infty}\left(\frac{n}{n+1}\right) e^{\log \log^2 n-\log \log^2 (n+1)}$$
How do i further simplify this limit ? Is my attempt correct or did i mess something up ? any suggestions are welcome !! 
These are exercise problems from Serge Lang's Undergraduate analysis, Chapter IX, exercise 6.
 A: 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.
A: $\lim\limits_{n \to \infty} \frac{n+1}{n} = 1$ 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 $1$.
A: 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.
