radius of convergence for $a_n= (\log n)^2$ determine the radius of convergence of the power series for:
$$a_n= (\log (n))^2$$
I know i am suppose to use the Ratio test and L'Hopitals Rule. I can't seem to figure out the steps 
 A: The radius of convergence is defined by the following equation:
$$r^{-1}=\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
We have, hence:
$$
\begin{align}
r^{-1}&=\lim_{n \to \infty}\left|\frac{(\log(n+1))^2}{(\log(n))^2}\right|\\
&=\lim_{n \to \infty}\left|\left(\frac{\log(n+1)}{\log(n)}\right)^2\right|\\
&=\lim_{n \to \infty} \left(\frac{\log(n+1)}{\log(n)}\right)^2\\
&=\lim_{n \to \infty} (\log_{n}(n+1))^2\\
&=1.\\
r^{-1}=1 &\Rightarrow r=1.
\end{align}
$$
Addendum
Here's how you apply Bernoulli's rule:
$$
\begin{align}
\lim_{n \to \infty}\left(\frac{\log(n+1)}{\log (n)}\right)^2&=\lim_{n \to \infty}\frac{((\log(n+1))^{2})^{\prime}}{((\log(n))^2)^{\prime}}.\\
\text{Given } (f \circ g)^{\prime}(n)&=(f^{\prime}\circ g)(n)g^{\prime}(n),\\
((\log(n+1))^{2})^{\prime}&=2(\log(n+1))(\log(n+1))^{\prime}.\\ (\log(n+1))^{\prime}&=\frac{1}{n+1}\cdot 1=\frac{1}{n+1}.\\
\text{Therefore, } ((\log(n+1))^{2})^{\prime}&=2(\log(n+1))\cdot \frac{1}{n+1}.\\
((\log(n))^2)^{\prime}&=2(\log n)\cdot \frac{1}{n}.\\
\text{Hence, } \lim_{n \to \infty}\frac{((\log(n+1))^{2})^{\prime}}{((\log(n))^2)^{\prime}}&=\lim_{n \to \infty}\frac{2(\log(n+1))\cdot \frac{1}{n+1}}{2(\log n)\cdot \frac{1}{n}}\\
&=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\frac{n}{n+1}\\
&=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\lim_{n \to \infty}\frac{n}{n+1}.\\
\lim_{n \to \infty}\frac{n}{n+1}&=1.\\
\text{Thus, } \lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\lim_{n \to \infty}\frac{n}{n+1}&=\lim_{n \to \infty}\frac{\log(n+1)}{\log(n)}\\
&=\lim_{n \to \infty}\frac{(\log(n+1))^{\prime}}{(\log(n))^{\prime}}\\
&=\lim_{n \to \infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}\\
&=\lim_{n \to \infty}\frac{n}{n+1}\\
&=1.
\end{align}
$$
A: The radius is clearly less than one, since $a_n$ tends to infinity. Conversely, for any $0<\rho <1$, the sequence $a_n \rho^n$ converges to zero. Therefore the radius of convergence is greater or equal to one, so it is one.
A: we know $\log n<n $, then (Hadamard formula):
$$\limsup ({{(\log n)}^\frac{2}{n}})<\limsup n^\frac{2}{n}=1.$$
Then the radius of convergence is 1.
