Loser66
  • Loser66
Find radius of convergence of \(\sum_{n=0}^\infty \dfrac{5^n}{n!} z^n\) . I got infinitive am I right? Please, help
Mathematics
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SOLVED
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katieb
  • katieb
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thomas5267
  • thomas5267
Is this the formula for the radius of convergence? \[ \limsup_{n\to\infty}\sqrt[n]{\frac{5^n}{n!}}z \]
thomas5267
  • thomas5267
Actually no.
freckles
  • freckles
what about ratio test

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thomas5267
  • 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
  • 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
  • freckles
like because there we would get |z|*0<1 which is true 0<1 for all z
freckles
  • freckles
oh so nevermind I think it works
thomas5267
  • thomas5267
But it only shows that it is absolutely convergent but not necessarily analytic over the Reals right?
freckles
  • 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
  • freckles
they do use root test there though instead of ratio test
Loser66
  • Loser66
Unfortunately, my Prof doesn't accept ratio test. We Must do Hamadard test.
Loser66
  • 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
  • Loser66
@misty1212
thomas5267
  • 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
  • thomas5267
I am not so sure about dropping the \(\sqrt{2\pi n}\) part of the Stirling's approximation.
Loser66
  • Loser66
But then how to argue for R?
Loser66
  • Loser66
R is a number, not limit anymore. 1/0 is undefined
Loser66
  • Loser66
Can we say the radius of convergence is infinitive?
thomas5267
  • 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
  • Loser66
Yes, I got you. I know that the radius of convergence is infinitive is different from divergence.
Loser66
  • Loser66
Thank you so much.

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