How to find the radius of convergence and the interval of convergence for a power series?

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.

$$\sum_{n=3}^{\infty} \frac{(x-1)^n}{n \sqrt{ln(n)}}$$

$\lim_{ n \to \infty} |\frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n}|$

= $\lim_{ n \to \infty}|\frac{(x-1)^{n+1}}{(n+1)\sqrt{ln(n+1)}} * \frac{n \sqrt{ln(n)}}{(x-1)^n}|$

= $\lim_{ n \to \infty}| \frac{(x-1)}{(n+1)\sqrt{ln(n+1)}} * n\sqrt{ln(n)}|$

• Why did it change from $x-2$ to $x-1$? – Andrew Li Apr 16 '18 at 0:14
• It was supposed to be x-1 in the question, ill fix that. – dg123 Apr 16 '18 at 0:19

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 $[0,2)$.
• So since the limit approaches 1, $|x-1|<1$? – dg123 Apr 16 '18 at 0:29
• It helps to know $n^{1/n}$ tends to 1. That automatically implies that things that increase slower than $n$ like $ln(n)$ also tend to 1 when raised to $1/n$. As for $n^{1/n}$ it's just $e^{\frac{1}{n}ln(n)}$, and you can see that the exponent tends to 0 (use L'Hospital's rule if you need to), so the whole expression tends to 1. – C Monsour Apr 16 '18 at 0:41