Series Convergence Test for 2 < x < 10: Value of X in a Series

[duki]

Homework Statement

Homework Equations

The Attempt at a Solution

I'm testing the end points of 2 < x < 10.
The Series looks like this:

SUM$$\frac{(x-6)^n}{4^n\sqrt{n}}$$

The first endpoint (10), I got diverges by the p-test (p <= 1). The second one, in my notes it says to take the lim, and if it's 0 then it converges by alt. series test. Why do I do two different tests?

 

[Dick]

You take two different tests because they are two different series. One converges (conditionally) and one doesn't. I don't see what the problem is.

 

[HallsofIvy]

At x= 10, You have
$$\sum \frac{4^n}{4^n\sqrt{n}}= \sum\frac{1}{n^{1/2}}$$
so the p test applies.

At x=1 you have
[tex]\sum \frac{(-5)^n}{4^n \sqrt{n}}= \sum \left(\frac{-5}{4}\right)^n\frac{1}{n^{1/2}}[/itex]<br /> which is an alternating series.[/tex]

 

[duki]

Ok, maybe I just need help with my exponents. Why did the n exponents cancel in the first one but not in the second one?

I got
$$\sum \frac{(-4)^n}{4^n \sqrt{n}}= \sum \left(\frac{-4}{4}\right)^n\frac{1}{n^{1/2}} = \frac{-1}{n^{1/2}}$$
for the second one.

 

[HallsofIvy]

Because 4/4= 1 and -5/4 does not!

 

[duki]

I edited my post. sorry