1. ## Radius and interval of convergence

Hi, I'm trying to do this question, but I'm stuck.

Find the radius and interval of convergence of the series:

x^2n/(n(ln(n))^2)

The summation from 2 to inifinity.

I applied the ratio test and got that the limit equals abs(1/x^2). Applying the inequality I got abs(1/x^2) < 1.

I'm stuck here, because shouldn't the radius of convergence be 1? According to the inequality, the series will converge on two intervals: [-inf, -1) and (1, inf].

Some help would be greatly appreciated.

2. Because $\displaystyle \forall n>1$ is $\displaystyle |\frac{1}{n^{2}\cdot \ln^{2} n}|<1$ and the series...

$\displaystyle \sum_{n=2}^{\infty} x^{2n}$ (1)

... converges for $\displaystyle |x|<1$, also the series...

$\displaystyle \sum_{n=2}^{\infty} \frac{x^{2n}}{n^{2}\cdot \ln^{2} n}$ (2)

... converges for $\displaystyle |x|<1$. Ther only difference is that the series (2), unlike the (1), converges also for $\displaystyle |x|=1$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$