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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.

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

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

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

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

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

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    $\begingroup$ By the way, once you are used to the root test, you won't ever use the ratio test unless the ratio test is obviously going to be easy. That's because the root test works whenever the ratio test does, but not conversely. $\endgroup$ – C Monsour Apr 16 '18 at 0:24
  • $\begingroup$ I agree with your point on the ratio test, but since I don't use the root test much, how did you find that the nth root of the denominator tends to 1? $\endgroup$ – Andrew Li Apr 16 '18 at 0:27
  • $\begingroup$ So since the limit approaches 1, $|x-1|<1$? $\endgroup$ – dg123 Apr 16 '18 at 0:29
  • $\begingroup$ 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. $\endgroup$ – C Monsour Apr 16 '18 at 0:41
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    $\begingroup$ Once you know the radius of convergence, those are the only two values you still need to test, yes. $\endgroup$ – C Monsour Apr 19 '18 at 23:48

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