# Homework Help: Power series

1. Nov 28, 2004

### kdinser

For the most part I seem to understand what I'm doing, but whenever you toss in a $$(-1)^n$$ or a $$(-1)^{n+1}$$ it starts to trip me up.
For example:

$$\sum\frac{(-1)^{n+1}(x-5)^n}{n5^n}$$

pretty straight forward, it's centered at 5, use the ratio test and solve the inequality to obtain R = 5 with endpoints 0, 10. It's when I go to test the endpoints that I get into trouble. I am going to solve this the way I would and maybe someone can point out where I'm going wrong with my thinking or algebra.

$$\sum\frac{(-1)^{n+1}(0-5)^n}{n5^n}$$

$$\sum\frac{(-1)^{n+1}(-5)^n}{n5^n}$$

Just as I was typing this, I think I found my mistake, but how do I fix it?
I have been factoring out a $$-1$$ from the $$(-5)^n$$ and canceling the $$5^n$$ with the one in the denominator. You can't do that can you? How about this?

$$\sum\frac{(-1)^{n+1}(-1)^n(5)^n}{n5^n}$$

That would let the 5^n's cancel and would leave

$$\sum\frac{(-1)^{2n+1}}{n}$$

right? Isn't this still an alternating series? The solutions manual says that when x=0 it should go to

$$\sum\frac{-1}{n}$$

which is a divergent p series and when x = 10 it stays an alternating series.

2. Nov 28, 2004

### marlon

You are right. When x=0 you get a divergent p series. But the $$\sum\frac{(-1)^{2n+1}}{n}$$ is NO alternating series because the sign always stays minus. If x = 10, you will get an alternating series : $$\sum\frac{(-1)^{n+1}}{n}$$

regards
marlon

3. Nov 28, 2004

### kdinser

Thanks marlon, I see it now. No matter what n is, your always going to get -1 raised to an odd value.