Loser66
  • Loser66
Find the radius of convergence of \(\sum_{n=1}^\infty \dfrac{n^{n+2}}{n!}z^n\) I got e. Am I right?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
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 this expert

answer on brainly

SEE EXPERT ANSWER

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

SolomonZelman
  • SolomonZelman
it seems as though it is |z|<1/e if you think of stirling's approximation z! = z^z • √(2πz) ÷ e^z
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
you got it upside down i think oh, what @SolomonZelman said

Looking for something else?

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

More answers

misty1212
  • misty1212
ratio test works right?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n+2}z^n}{n!} }\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty }\frac{n^{n}n^2z^n}{n!} }\) 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.
SolomonZelman
  • SolomonZelman
do the ratio test tho, that might be off just a bit.
misty1212
  • misty1212
\[\frac{(n+1)^{n+3}\times n!}{(n+1)!n^{n+2}}=\frac{(n+1)^{n+2}}{n^{n+2}}\]
misty1212
  • misty1212
you do get \(e\) as the limit, but that means \[|z|<\frac{1}{e}\]
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \lim_{n \to \infty} \frac{n!(n+1)^{n+3}z^{n+1}}{(n+1)!n^{n+2}z^n} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+3}}{(n+1)!n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{n!(n+1)^{n+2}}{n!\cdot n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\frac{(n+1)^{n+2}}{n^{n+2}} }\) \(\large\color{black}{ \displaystyle \lim_{n \to \infty} z\left(1+\frac{1}{n}\right)^{n+2}=ez}\) |ez|<1 |z|<1/e
SolomonZelman
  • SolomonZelman
same thing:)
Loser66
  • Loser66
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.
SolomonZelman
  • SolomonZelman
Anytime....
SolomonZelman
  • SolomonZelman
Yeah, just a note that this denominator of ≈n/e is also a consequence of Stirling's approximation.
SolomonZelman
  • SolomonZelman
in any case, it is *|z|<1/e* .....

Looking for something else?

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