## Loser66 one year ago Find radius of convergence of $$\sum_{n=0}^\infty \dfrac{5^n}{n!} z^n$$ . I got infinitive am I right? Please, help

1. thomas5267

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

2. thomas5267

Actually no.

3. freckles

4. thomas5267

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.

5. freckles

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

6. freckles

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

7. freckles

oh so nevermind I think it works

8. thomas5267

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

9. freckles

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....

10. freckles

they do use root test there though instead of ratio test

11. Loser66

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

12. Loser66

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

13. Loser66

@misty1212

14. thomas5267

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

15. thomas5267

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

16. Loser66

But then how to argue for R?

17. Loser66

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

18. Loser66

Can we say the radius of convergence is infinitive?

19. thomas5267

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

20. Loser66

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

21. Loser66

Thank you so much.