# 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

• I hope my addendum has helped elucidate the more detailed matters of this question.
– 000
Oct 21, 2012 at 0:26
• How would one do this using the Hadamard formula? Jan 3, 2022 at 16:22

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}

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}
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.
we know $\log n<n$, then (Hadamard formula):
$$\limsup ({{(\log n)}^\frac{2}{n}})<\limsup n^\frac{2}{n}=1.$$