- #1
kdinser
- 337
- 2
For the most part I seem to understand what I'm doing, but whenever you toss in a [tex](-1)^n[/tex] or a [tex](-1)^{n+1} [/tex] it starts to trip me up.
For example:
[tex]\sum\frac{(-1)^{n+1}(x-5)^n}{n5^n}[/tex]
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.
[tex]\sum\frac{(-1)^{n+1}(0-5)^n}{n5^n}[/tex]
[tex]\sum\frac{(-1)^{n+1}(-5)^n}{n5^n}[/tex]
Just as I was typing this, I think I found my mistake, but how do I fix it?
I have been factoring out a [tex]-1[/tex] from the [tex](-5)^n[/tex] and canceling the [tex]5^n[/tex] with the one in the denominator. You can't do that can you? How about this?
[tex]\sum\frac{(-1)^{n+1}(-1)^n(5)^n}{n5^n}[/tex]
That would let the 5^n's cancel and would leave
[tex]\sum\frac{(-1)^{2n+1}}{n}[/tex]
right? Isn't this still an alternating series? The solutions manual says that when x=0 it should go to
[tex]\sum\frac{-1}{n}[/tex]
which is a divergent p series and when x = 10 it stays an alternating series.
For example:
[tex]\sum\frac{(-1)^{n+1}(x-5)^n}{n5^n}[/tex]
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.
[tex]\sum\frac{(-1)^{n+1}(0-5)^n}{n5^n}[/tex]
[tex]\sum\frac{(-1)^{n+1}(-5)^n}{n5^n}[/tex]
Just as I was typing this, I think I found my mistake, but how do I fix it?
I have been factoring out a [tex]-1[/tex] from the [tex](-5)^n[/tex] and canceling the [tex]5^n[/tex] with the one in the denominator. You can't do that can you? How about this?
[tex]\sum\frac{(-1)^{n+1}(-1)^n(5)^n}{n5^n}[/tex]
That would let the 5^n's cancel and would leave
[tex]\sum\frac{(-1)^{2n+1}}{n}[/tex]
right? Isn't this still an alternating series? The solutions manual says that when x=0 it should go to
[tex]\sum\frac{-1}{n}[/tex]
which is a divergent p series and when x = 10 it stays an alternating series.