Domain of convergence of power series-2 Here is the series:
$$
\sum_{n=2}^\infty(-1)^n\frac{(x-3)^n}{(\sqrt[n]{n}-1)n}
$$
What is the interval of convergence?
I tried using root test and ratio test but finding the limit from thereon is quite difficult.
From root test: 
$$
\lim_{n->\infty}\frac{|x-3|}{(\sqrt[n]{\sqrt[n]{n}-1)n}}
$$
 A: Let's go for the root test:
$$
r=\displaystyle\lim_{n\to\infty}\frac{|(x-3)|}{\sqrt[n]{(\sqrt[n]{n}-1)n}}
$$
So we have to calculate the following limit:
$$
\lim_{n\to\infty}{\sqrt[n]{(\sqrt[n]{n}-1)n}}
$$
Form here on, there is a mundane way to show that the limit is one. The direct path is using multiple Hopital. But we give another proof:
$$
\lim_{n\to\infty}{\sqrt[n]{(\sqrt[n]{n}-1)n}}=\lim_{n\to\infty}{\sqrt[n]{n}}{\sqrt[n]{\sqrt[n]{n}-1}}
$$
It can be shown that ${\sqrt[n]{n}}$ goes to 1 as $n\to\infty$. For the other limit, see that:
$$
\frac{\ln n}{n}\leq \sqrt[n]{n}-1\leq n\implies  \sqrt[n]{\frac{\ln n}{n}}\leq \sqrt[n]{\sqrt[n]{n}-1}\leq \sqrt[n]{n}
$$
Now you can show that both limits of LHS and RHS go to 1 as $n\to\infty$ and therefore the limit in the middle exists and is equal to one and hence the final limit exists and is 1. 
In any case the domain of convergence is:
$$
{|(x-3)|}<1
$$
with the point $x=4$.

Remark 1:
$$
\sqrt[n]{\frac{\ln n}{n}}=L\implies \frac{1}{n}\ln(\frac{\ln n}{n}) =\ln L\\
\lim_{n\to\infty} \frac{\ln({\ln n})-\ln({n})}{n}=0\implies L=1.
$$
Remark 2: If $x-3=-1$, the series becomes:
$$
\sum_{n=2}^\infty(-1)^{2n}\frac{1}{(\sqrt[n]{n}-1)n}=\sum_{n=2}^\infty\frac{1}{(\sqrt[n]{n}-1)n}
$$
Such series with positive non-increasing terms converges iff $\sum 2^na_{2^n}$ converges. Hence:
$$
\sum_{n=2}^\infty\frac{2^n}{(\sqrt[n]{2^n}-1)2^n}=\sum_{n=2}^\infty {1}\to\infty
$$
Therefore it diverges. 
For the case were $x-3=1$, we get an alternating series with non-increasing absolute value of its terms and according to the alternating series test it converges.
