power series

[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.

[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

[kdinser]

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