$\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{nln(1+n)}$

I have attempted this problem as following

By root test

$\displaystyle lim_{n\to\infty}\sqrt[n]{|\frac{x^n}{nln(1+n)}|}$

$\displaystyle =lim_{n\to\infty}|x|\sqrt[n]{|\frac{1}{nln(1+n)}|}$

$\displaystyle =lim_{n\to\infty}|x|\frac{1}{|\sqrt[n]{ln(1+n)}|}$

$\displaystyle =\frac{|x|}{lim_{n\to\infty}\sqrt[n]{ln(1+n)}}$

$\displaystyle =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}$

$\displaystyle =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}$

$\displaystyle =\frac{|x|}{\sqrt[n]{ln(lim_{n\to\infty}(1+n))}}$

$\displaystyle =|x|$

When $\displaystyle x=1$,$\displaystyle \sum_{n=0}^{\infty}\frac{1}{nln(1+n)}$ converges.

When $\displaystyle x=-1$,$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{nln(1+n)}$converges.

Therefore, $\displaystyle R=1$ and $\displaystyle I=[-1,1]$

Is it correct?

#2

Assume that the positive series $\displaystyle \sum_{n=0}^{\infty} a_n$ is convergent.

Prove that the series $\displaystyle \sum_{n=0}^{\infty} ({a_n}^4+\frac{1}{n^2})$ is also convergent.