Question

Can someone that is comfortable explaining infinite series please help?????

Answer 1

how does
\[\sum_{n=0}^{\infty}n!x ^{n}=1+0+0+0....=1\]

Answer 2

This is for x=0. I know that 0!=1 but how does that apply to the first term? you have 1*0^0 so then how is that one?????

Answer 3

Appearently 0^0=1.
Hmm I had never really thought about that before. D:
Seems unintuitive. I'll have to read up on that.

Answer 4

o^0 is not equal to one, it is an indeterninate form as far as I have been told

Answer 5

Are you sure that n factorial doesn't go to the denominator and that the whole thing isn't multiplied by \(e^{-x}\). I say this because:
$$
\bf
\sum_{n-0}^{\infty}\dfrac{x^n}{n!}=e^x
$$

Answer 6

I am one hundred percent sure, I have to learn power series on my own because of the differences in course outlines from one school to the next. This example comes directly from the textbook

Answer 7

And this is for x=0 or for x near 0? :o just checking.

Answer 8

it uses this example with the ratio test to show that the limit is = infinity. Then they say "therefore the power series converges only at it's center with a radius of convergence=0" that makes zero sense to me. I thought the ratio test said that if the limit was greater than z4ero the series diverges. But this one is equal to infinity and converges at x=0? doesn't make sense

Answer 9

@zepdrix I am going to post verbatim what it says give me one second

Answer 10

The only logical conclusion is that your book assumes that 0^0 =1. But this is by no means definitive:
https://cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node14.html

Answer 11

You are correct, by the ratio test this series diverges because the ratio is greater than 1.

Answer 12

Find the radius of convergence of \[\sum_{n=0}^{\infty}n!x ^{n}\]
Solution:
For x=0 you obtain\[f(0)=\sum_{n=0}^{\infty}n!0^{n}=1+0+0+0....=1\]
for any fixed value such that\[\left| x \right|>0\] let u=\[n!x ^{n}\] then \[\lim_{n \rightarrow \infty}\left| \frac{ u _{n+1} }{ u _{n} } \right|=\left| x \right|\lim_{n \rightarrow \infty}(n+1)=\]Infinity
Therefore by the ratio test the series diverges for\[\left| x \right|>0\]

Answer 13

and converges only at its center,0. So the radius of convergence is r=0

Answer 14

And that folks is what has me stumped

Answer 15

all of it makes sense accept for the part where the series equals one when x=0

Answer 16

Yah it looks like they're assuming 0^0 to be 1.
That's the only way that I can see \(\large f(0)\) equalling \(\large 1\).

Answer 17

The rest of it makes sense though? Oh that's good :)

Answer 18

That's right. The radius of convergence can also e determined by :

$$
\Large \frak{R}\tt=\lim_{n\to\infty} \dfrac{n!x^n}{(n+1)!x^{n+1}}=0
$$

Answer 19

@zepdrix I agree but I don't like the 0^0 BECAUSE THIS SAME BOOK Teaches that as an indeterminate form 'From which yo can make no conlusions" ane @ybarrap I don't recognize that

Answer 20

Hmm yah that's weird :(

Answer 21

You have to assume that the author borrowed someone else's book that recognizes 0^0=1 and so therefore is a typo in consistency.

Answer 22

@SithsAndGiggles You are very good at these do you have anything to add to this?

Answer 23

Thank you all for your help Hopefully one day I can help you guys/gals with something (not likely cause you are all GURU's)

Answer 24

One final word from the Maestro, Wolfram: http://www.wolframalpha.com/input/?i=0%5E0

Answer 25

I checked that before I even waisted anyone's time @ybarrap lol

Answer 26

Yeah I do that, too

Answer 27

Nope, nothing to add, I share the same concerns.

Answer 28

Ok thanks again everyone