Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- Loser66

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- Loser66

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- thomas5267

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

- thomas5267

Actually no.

- freckles

what about ratio test

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- 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.

- 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

- freckles

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

- freckles

oh so nevermind I think it works

- thomas5267

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

- 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....

- freckles

they do use root test there though instead of ratio test

- Loser66

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

- 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.

- Loser66

- 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?

- thomas5267

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

- Loser66

But then how to argue for R?

- Loser66

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

- Loser66

Can we say the radius of convergence is infinitive?

- 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

- Loser66

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

- Loser66

Thank you so much.

Looking for something else?

Not the answer you are looking for? Search for more explanations.