What is radius of convergence of series $\sum_{n=0}^{\infty} ({\log n})^2 (z^{n})$
I know that for a holomorphic function $f$ whose power series has coefficient $a_n$ is given as
$\frac{1}{R}= \lim_{x \to \infty} |\frac{a_{n+1}}{a_n}|$
Using this I am stuck in the step $\frac{1}{R}= \lim_{x \to \infty} |\frac{\log({n+1})}{\log n}|$ , Please help.
Your expression for $\frac{1}{R}$ in the last line is wrong — why is $a_n$ still there? There's the general formula, but when you use it for a specific series, both $a_n$ and $a_{n+1}$ must be replaced by the corresponding expressions from the given series.
In this example: $a_n=(\log n)^2$ and $a_{n+1}=(\log(n+1))^2$. (And I'm presuming $\log$ stands for the natural logarithm, although that doesn't really matter.) So $$\frac{1}{R}=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{(\log(n+1))^2}{(\log n)^2}\right|=1,$$ where the limit can be found e.g. using L'Hospital's Rule.
The Radius of Convergence should be 1. The problem with your last line is that you are putting ${a_n}$ again with ${log}$ term .The log term is itself your${ a_n}$