1. Dec 6, 2017

### WWGD

Nice plug! EDIT: You're beating me at this, you should be worried :).

2. Dec 6, 2017

### FactChecker

I like and agree with your edit comment.

3. Dec 6, 2017

### ScreamingIntoTheVoid

Right, that makes sense, thank you for taking the time to provide me with a more extensive responce. And r would be found using the ratio test correct? So just to make sure that I'm doing this right, if I had my original problem (except written correctly):

Σ (x-2)n/nn , using the ratio test I would get (x-2)n+1/nn+1 * (nn)/(x-2)n which I
n→1
could turn into [(x-2)(x-2)n] * (nn)/(x-2)n , which by applying a limit and simplifying that would turn into:
|x-2| lim n→∞ (nn/(n+1)n+1) which would turn into |x-2|>0, leaving the interval of convergence to be -2>n>2 and the r value to be 2?

(note: That's probably a somewhat messy version of the process if I did it right because I also taught myself intervals of convergence/the ratio of convergence last night. Though I feel like I have somewhat of a grasp on those, I've probably done it in a somewhat messy manner)

**ALSO thanks for responding to this thread even though it's just about a day old**

Last edited by a moderator: Dec 6, 2017
4. Dec 7, 2017

### Ray Vickson

Sometimes the ratio test can be used to determine the radius of convergence, but sometimes other tests must be used. Sometimes we really do not need any such tests at all, but can just rely on a bounding property.

The radius of convergence of $\sum (x-2)^n/n^n$ is $\infty$, not 2 or 3 or any other finite number.

First: simplify writing by putting $x-2 = y$, so the series is $\sum y^n/n^n$. If we set $t_n = y^n/n^n$ we have the ratio
$$\frac{|t_{n+1}|}{|t_n|} = \frac{|y|^{n+1}}{|y|^n} \frac{n^n}{(n+1)^{n+1}} = |y| \left( \frac{n}{n+1}\right)^n \frac{1}{n+1}.$$
Now $n/(n+1) < 1$ so $(n/(n+1))^n < 1$ for all $n > 0$, so we have
$$\frac{|t_{n+1}|}{|t_n|} < |y| \, \,\frac{1}{n+1} \to 0\; \text{as} \; n \to \infty.$$
The limiting ratio is <1 for any $|y|$---and, in fact, is < 0.1 or < 0.000000000001 or .... or less than any positive number at all. Therefore, the sum will converge, no matter what is the value of $y = x-2$, so will converge no matter what is the value of $x$. It converges for $x = -1,000,000$ or for $x = 175,000$ or whatever other value you choose to employ.

The radius of convergence cannot be 2 because you would be saying that the series diverges if $|x-2| > 2$, and that is definitely not the case here.

Last edited: Dec 7, 2017