anonymous
  • anonymous
Limits of series. I am going to type the equation using equation editor.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I am trying to find the following limit. \[\lim_{n \rightarrow \inf} \frac{ 2(n!) }{ (-6)^n }\]
SolomonZelman
  • SolomonZelman
Factorial will grow much much bigger than any exponential growth (after some point).
anonymous
  • anonymous
Is it because the factorial is multiplied by 2?

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anonymous
  • anonymous
I was under the assumption that the (-6)^n would grow faster than the factorial.
SolomonZelman
  • SolomonZelman
two does not do anything at all.... whether it has a two or not factorial is much bigger. Wrong assumption! The opposite is true
anonymous
  • anonymous
Then I guess the limit diverges to both negative and positive infinity.
SolomonZelman
  • SolomonZelman
Yes:) positive and negative infinity, because it alternates.
anonymous
  • anonymous
Alright, Thanks. I guess my understand of factorials are a bit off.
SolomonZelman
  • SolomonZelman
We can play around and maybe apply Stirling's approximation: (Note, that the factorial is greater than its stirlings approximation) \(\large\color{black}{ \displaystyle n!\approx\frac{\sqrt{2\pi n}{~}n^n}{e^n} }\) \(\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\frac{\sqrt{2\pi n}{~}n^n}{6^ne^n} }\) \(\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\sqrt{2\pi n}\frac{{~}n^n}{(6e)^n} }\)
some.random.cool.kid
  • some.random.cool.kid
thx i learned something new just by watching lol
SolomonZelman
  • SolomonZelman
You would then agree that the numerator is greater, right?
anonymous
  • anonymous
I do agree that the numerator is greater.
SolomonZelman
  • SolomonZelman
Good:) And I understand that you were applying the Diverghence test to \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{2(n!)}{(-6)^n}}\)
SolomonZelman
  • SolomonZelman
(not necessarily n=1, but that series)
anonymous
  • anonymous
yep
SolomonZelman
  • SolomonZelman
Ok, so if it was the other way around, then you would apply: \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{x^n}{n!}=e^x}\) So for example: \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{(-6)^n}{n!}=e^{-6}= 1/e^6}\)
SolomonZelman
  • SolomonZelman
Anyway, good luck....
anonymous
  • anonymous
Thanks for the help. It really made since. I am now making forward progress on my assignments.
SolomonZelman
  • SolomonZelman
Good to hear that:)
SolomonZelman
  • SolomonZelman
be well

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