Question

Find radius of convergence of
\(\sum_{n=0}^\infty \dfrac{5^n}{n!} z^n\) . I got infinitive am I right?
Please, help

Answer 1

Is this the formula for the radius of convergence?
\[
\limsup_{n\to\infty}\sqrt[n]{\frac{5^n}{n!}}z
\]

Answer 2

Actually no.

Answer 3

what about ratio test

Answer 4

I thought he wanted to show that the radius of convergence is infinite. I have not taken any analysis course but Wikipedia says that ratio test only shows that the radius of convergence is finite.

Answer 5

\[n \rightarrow \infty \\ |\frac{a_{n+1}}{a_n}|<1 \\ |\frac{5^{n+1}}{(n+1)!}z^{n+1} \cdot \frac{n!}{5^{n} } \frac{1}{z^n}|<1 \\ |\frac{5}{(n+1)} z|<1 \\ | z| |\frac{5}{n+1}|<1 \\ \text{ remember } n \rightarrow \infty \]
maybe you are right

Answer 6

like because there we would get |z|*0<1
which is true 0<1 for all z

Answer 7

oh so nevermind I think it works

Answer 8

But it only shows that it is absolutely convergent but not necessarily analytic over the Reals right?

Answer 9

I don't know about all of that.
I just know it shows it converges for all z since we have |z|*0<1 is true for all z
therefore the radius of convergence is infinite

there is a similar example here: http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx
in example 4....

Answer 10

they do use root test there though instead of ratio test

Answer 11

Unfortunately, my Prof doesn't accept ratio test. We Must do Hamadard test.

Answer 12

\(limsup \sqrt[n]\dfrac{5^n}{n!} = limsup \dfrac{5}{n/e} = limsup \dfrac{5e}{n} = \infty \)
Hence \(R = \dfrac {1}{\infty}= ?\) I don't know how to argue then, since it is undefined.

Answer 13

@misty1212

Answer 14

\[
\limsup_{n\to\infty} \sqrt[n]{\dfrac{5^n}{n!}} = \limsup_{n\to\infty} \dfrac{5}{n/e} = \limsup_{n\to\infty} \dfrac{5e}{n} = 0
\]
Since n is to infinity right?

Answer 15

I am not so sure about dropping the \(\sqrt{2\pi n}\) part of the Stirling's approximation.

Answer 16

But then how to argue for R?

Answer 17

R is a number, not limit anymore. 1/0 is undefined

Answer 18

Can we say the radius of convergence is infinitive?

Answer 19

If the sequence values are unbounded so that the lim sup is ∞, then the power series does not converge near a, while if the lim sup is 0 then the radius of convergence is ∞, meaning that the series converges on the entire plane.

https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem

Answer 20

Yes, I got you. I know that the radius of convergence is infinitive is different from divergence.

Answer 21

Thank you so much.