A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Loser66

  • one year ago

Find the radius of convergence of \(\sum_{n=1}^\infty \dfrac{n^{n+2}}{n!}z^n\) I got e. Am I right?

  • This Question is Closed
  1. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it seems as though it is |z|<1/e if you think of stirling's approximation z! = z^z • √(2πz) ÷ e^z

  2. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    HI!!

  3. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you got it upside down i think oh, what @SolomonZelman said

  4. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ratio test works right?

  5. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n+2}z^n}{n!} }\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2z^n}{n!} }\) by Stirling's approximation, \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2e^nz^n}{n^n\sqrt{2\pi n}} }\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^2e^nz^n}{\sqrt{2\pi n}} }\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^2e^n}{\sqrt{2\pi n}\left(\dfrac{1}{z}\right)^n} }\) So if: |1/z|>e ---> z<1/e then the series will converge.

  6. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do the ratio test tho, that might be off just a bit.

  7. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{(n+1)^{n+3}\times n!}{(n+1)!n^{n+2}}=\frac{(n+1)^{n+2}}{n^{n+2}}\]

  8. misty1212
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you do get \(e\) as the limit, but that means \[|z|<\frac{1}{e}\]

  9. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\large\color{black}{ \displaystyle \lim_{n \to \infty} \frac{n!(n+1)^{n+3}z^{n+1}}{(n+1)!n^{n+2}z^n} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+3}}{(n+1)!n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+2}}{n!\cdot n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{(n+1)^{n+2}}{n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\left(1+\frac{1}{n}\right)^{n+2}=ez}\) |ez|<1 |z|<1/e

  10. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    same thing:)

  11. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I use Hamadard test, which is \(a_n = \dfrac{n^{n+2}}{n!}\) then \(\sqrt[n]|a_n| = \dfrac{\sqrt[n]{n^2n^n}}{\sqrt[n]n!}\) and we know that the denominator \(\approx n/e \). Hence, it is \(\dfrac{n\sqrt[n] n^2}{(n/e)}= e\sqrt[n]n^2\) oh, yeah, I forgot to take \(R = 1/ limsup \sqrt[n]|a_n|\) that is why I got e, but 1/e Thank you all.

  12. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Anytime....

  13. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah, just a note that this denominator of ≈n/e is also a consequence of Stirling's approximation.

  14. SolomonZelman
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    in any case, it is *|z|<1/e* .....

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.