For this question, I'm stuck on finding the radius of convergence and interval of convergence for the power series. Here is what I have so far. Can anyone please help me out?
Find the radius of convergence and the interval of convergence for the power series.
$$\sum_{n=3}^{\infty} \frac{(x-1)^n}{n \sqrt{ln(n)}}$$
$\lim_{ n \to \infty} |\frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n}|$
= $\lim_{ n \to \infty}|\frac{(x-1)^{n+1}}{(n+1)\sqrt{ln(n+1)}} * \frac{n \sqrt{ln(n)}}{(x-1)^n}|$
= $\lim_{ n \to \infty}| \frac{(x-1)}{(n+1)\sqrt{ln(n+1)}} * n\sqrt{ln(n)}|$
The radius of convergence is easy: It's 1. Just use the $n^{th}$ root test instead of the ratio test. The $n^{th}$ root of the denominator tends to 1 and the $n^{th}$ root of the absolute value of the numerator goes to $|x-1|$, which is less than 1 for $x\in(0,2)$ and greater than 1 for $x\notin[0,2]$. Then you still need to figure out what happens at x=0 and x=2. For x=2 this simplifies to a series that is well known to diverge, as you can see by doing the approximating integral with by substitution of $u$ for $ln(n)$. For x=0 you have an alternating series with decreasing terms, so it converges, so the interval of convergence is $[0,2)$.