# Find the Interval of Convergence

1. Apr 4, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
Find the interval of convergence.

2. Relevant equations
$\displaystyle \sum^{∞}_{n=0} \frac{(x-5)^n}{n^4 * 2^n}$

3. The attempt at a solution
I used the ratio test as follows:

$\displaystyle \frac{(x-5)^{n+1}}{(n+1)^4 * 2^{n+1}} * \frac{n^4 2^n}{(x-5)^n}$

taking the limit:

(x-5) lim (x->inf) $\displaystyle \frac{n^4}{2(n+1)^4}$ = 0

Most of these problems the limit comes to 1. How do i solve this now that the limit is zero?

2. Apr 4, 2013

### CompuChip

You are very much on the right track, except that
$$\lim_{n \to \infty} \frac{n^4}{(n + 1)^4} \neq 0$$

You can see this if you expand $(n + 1)^4 = n^4 + \mathcal O(n^3)$

(Don't forget the factor of 1/2 in the end)

3. Apr 4, 2013

### whatlifeforme

|x-5| < 1

-1 < x-5 < 1
4 < x < 6

Answer (interval of convergence): 4 < x < 6

Does that look right?

4. Apr 4, 2013

### whatlifeforme

i think i forgot the factor of (1/2), right?

5. Apr 4, 2013

### HallsofIvy

Staff Emeritus
Yes, you did. The correct limit is 1/2. Now, what does that tell you about the radius of convergence?

6. Apr 4, 2013

### whatlifeforme

what is the radius of convergence vs interval of convergence?

7. Apr 4, 2013

### whatlifeforme

is this part correct? because that is one of the choices as the answer but 2 < x < 3 is not.

8. Apr 4, 2013

bump.

9. Apr 4, 2013

### CompuChip

The radius of convergence is the "width" of the interval of convergence. E.g. if you are looking at the series around x = 2, if the radius of convergence is 4 the interval of convergence would be 2 - 4 < x < 2 + 4, i.e. -2 < x < 6.

First try answering Halls' question. The radius of convergence is related to the ratio (1/2) you calculated.

Please don't do that, we are not online 24 hours per day and we will get back to you when we have time.

10. Apr 4, 2013

### whatlifeforme

sorry for the bump. i'm getting 3 < x < 7 now that I added in the factor of 1/2

-1 < (x-5)/2 < 1