1. Dec 14, 2004

### NINHARDCOREFAN

I don't understand how to get the inteveral of convergence from a problem:

for example:

infinity
_
\
/ -(1)^n(x-2)^n/(n*4^n)
_
n=1

I know you have to use ratio test and it comes out to:
|x-2|/4

The textbook says the series converges when |x-2|<4. How did they come to this conclusion? How do you find c? the answer was c=2. The series converges when -2<x<6 and diverges if either x<-2 or x>6. Can someone explain this also? Thanks

2. Dec 14, 2004

### fourier jr

$$\sum_{n=1}^\infty \frac{((-1)^n)(x-2)^n}{n4^n}$$ converges when a_n < 1 , so |x-2|/4 < 1 => |x-2|< 4.

|x-2| < 4 means that for -2 < x < 6, the series converges. plug x=-2 into the series & see that the series converges @ x=-2. plug x=6 into the series & it converges there also. so the interval of convergence is [-2, 6]

3. Dec 14, 2004

### NINHARDCOREFAN

hmm....

You just listed the facts and didn't explain them explicitly... anyway I understood most of the stuff through doing the problems. But I still don't understand one thing, when you plug in the x how do you find the limit when n is at infinity, I'm getting confused because of n being the power...

When you plugged in x=2 how do you solve for it?

4. Dec 14, 2004

### Tom Mattson

Staff Emeritus
Right.

They came to it by recognizing that under the ratio test a series converges absolutely if the limit as n goes to infinity of |an+1/an| is less than one. So you get:

|x-2|/4<1

or...

|x-2|<4

When you work the inequality above into the form:

|x-c|<R,

then c is the center of the series and R is the radius of convergence.

Use the definition of the absolute value.

|x-2|<4
-4<(x-2)<4
-2<x<6

Now that doesnt' tell you the endpoint convergence. You still have to test for that. So you plug x=-2 and x=6 into your power series (one at a time, of course) and you see if the resultant infinite series converge or diverge. That will tell you which inequality to put an "equal" sign under, if any.

5. Dec 14, 2004

### NINHARDCOREFAN

Thanks. But you didn't answer my main question, how do I solve the series after plugging in the x value (into the series)?

6. Dec 14, 2004

### Tom Mattson

Staff Emeritus
Have you not studied infinite series??

If you plug in x=-2, you get:

$$\sum_{n=1}^\infty \frac{((-1)^n)(-2-2)^n}{n4^n}$$
$$\sum_{n=1}^\infty \frac{((-1)^n)(-4)^n}{n4^n}$$
$$\sum_{n=1}^\infty \frac{((-1)^n)(-1)^n4^n}{n4^n}$$
$$\sum_{n=1}^\infty \frac{1}{n}$$

...which diverges by the integral test.

I'll leave it to you to plug in and test the series at x=6.

7. Dec 14, 2004

### fourier jr

doh! it's (x-2)^n not (x-2). just ignore my stuff...

8. Dec 15, 2004

### NateTG

There isn't necessarily an expression for the result of a sum. People are probably working on a nice expression for
$$\sum_{i=1}^{\infty} \frac{1}{n^3}$$
right now.

All this problem is asking for is convergence, so:

The ratio between the $n^{th}$ and $(n+1)^{th}$ terms is
$$\frac{n+1}{n} \frac{(-1)(x-2)}{4}$$
in the limit that turns into
$$\frac{(-1)(x-2)}{4}$$
and the ratio must be less than one for the series to converge, so
$$\left|\frac{(-1)(x-2)}{4}\right| < 1$$
$$\left|(x-2)\right| < 4$$
$$-2 < x < 6$$