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
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SOLVED
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jamiebookeater
  • jamiebookeater
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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

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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* .....

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