# 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.

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.

• What was i even doing? Thanks !! – Zeno San Oct 22 at 8:37
• I'm glad I could help. – José Carlos Santos Oct 22 at 8:38

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

• +1, this looks more elegant. – Zeno San Oct 22 at 8:38