# Finding an Interval of Convergence

1. Sep 2, 2008

### Battlemage!

1. The problem statement, all variables and given/known data

Find the interval of convergence of the infinite series:

(x-1)n / 2n
n = 1

2. Relevant equations

Using the ration test. It converges if the absolute value of the limit of f(n+1)/f(n) as n -> ∞ < 1.

Well, I hope that's how you write it. I'm sure you guys know how to use the ration test.

3. The attempt at a solution

I actually believe I have the answer. But I have no way of knowing if it is right.

Here is my answer:

The interval of convergence is -1 < x < 3.

I got this by using the ratio test, and then eliminating factors of (x-1)n and 2n (after expanding the exponents), leaving me with the limit of

lim (x-1)/2 as n -> ∞

Then, since there is no n, then that limit IS (x -1)/2 (unless I'm totally off)

And the absolute value of that has to be less than 1 for it to converge, so

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

-2 < x - 1 < 2

-1 < x < 3

Is this correct? Do you need me to post the part where I canceled out terms before I took the limit?

Thanks!

2. Sep 2, 2008

Looks good. The only thing the test doesn't tell you about is what happens at the endpoints ($$x = -1, x = 3$$). You have to try each one and determine what happens. (In general) You may find the series does not converge at either endpoint, it may converge at one but not the other, or it may converge at both.
Good work.

3. Sep 2, 2008

### Battlemage!

Thanks for the help, but I do have one more question. You see, it's been about 1.5 years since I've worked with series. This will be a very stupid question..

How exactly do you test if the two endpoints are included in the interval? I just don't know the method for doing that.

Thanks again!

4. Sep 2, 2008

Your most recent post begins:

Nonsense.
It continues

Good, common question. As an example, suppose I have the series

$$\sum_{n=1}^\infty \frac{x^n}{n}$$

Work similar to what you did in your example shows the radius of convergence to be $$1$$, so I know the series converges for $$-1 < x < 1$$. What about the endpoints here?
First, consider $$x = 1$$. Simply plug this value into the series and study the result. I get

$$\sum_{n=1}^\infty \frac{(1)^n}{n} = \sum_{n=1}^\infty \frac 1 n,$$

which I know diverges because it is the harmonic series. If I try $$= -1$$,

$$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$

which is known to converge.

Putting all of this work together shows that the original series converges for $$-1 \le x < 1$$, or $$[-1, 1)$$ if you write it in interval notation.

I hope this helps.