Question

Find the radiance of convergence.
sum_{n=1}^{infty} e^{-n^2}x^{n}

is the radius 0? I used the root test and the one of the limit goes to 0..

Answer 1

\[\sum_{n=1}^{\infty} e^{-n^2}x^{n}\]

Answer 2

the radius of convergence of the infinite series is 1 by the root test.

Answer 3

do you understand?

Answer 4

abs(e^-n x)<1 then absx<inf R=inf

Answer 5

for all x converge

Answer 6

\[ \sum_{n=1}^{\infty} e^{-n^2}x^{n} \]By ratio test: \[ \frac{e^{-(n+1)^2x^{n+1}}}{x^ne^{-n^2}} = x \frac{e^{-(n+1)^2}}{e^{-n^2}} = 0\]So R = infinity, no?

Answer 7

you shold test root thats beter i did

Answer 8

your prove is said r=inf also

Answer 9

so can I assume that whenever the limit goes to 0, R = infinity?

Answer 10

I agree, but both yield the same result, after computing the limit, we will reach 0 < 1, which is true and shows that R = infinity. And, yeah, I was writing, didn't see your post ;-)

Answer 11

@Dumboy You should think it like this: We got 0 and we want that the limit < 1 in order for the series to converge. For which values of x is 0 < 1? All of them, so R = infinity. So, yeah, if the limit goes to 0, you will get your answer as R = inf.

Answer 12

it means for all real x you have 0<1 a true

Answer 13

mahmit had it correct in his first post, and e^-n will go to 1 as x goes to infinity, so we get abs value of x is less than 1, so the radius is 1.

Answer 14

there is no limit for x cause x is disappeared

Answer 15

how did x disappear?

Answer 16

dont worry just believe!

Answer 17

using the root test leaves the abs. value of x is less than 1

Answer 18

yeas are the same limit

Answer 19

but the radius of convergence can't be infinity unless we do the limiting process and end up with a number that is less than 1, but we don't, we get x, correct?

Answer 20

No, we get 0 :-)

Answer 21

for all x correct means x belonging to all real

Answer 22

and we do get 0 and 0,1 so r = infinity, ok

Answer 23

ok E.x<1 then x<1/E and E go to 0 so x<inf

Answer 24

R = inf