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anonymous
 one year ago
Limits of series. I am going to type the equation using equation editor.
anonymous
 one year ago
Limits of series. I am going to type the equation using equation editor.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to find the following limit. \[\lim_{n \rightarrow \inf} \frac{ 2(n!) }{ (6)^n }\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Factorial will grow much much bigger than any exponential growth (after some point).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it because the factorial is multiplied by 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was under the assumption that the (6)^n would grow faster than the factorial.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3two does not do anything at all.... whether it has a two or not factorial is much bigger. Wrong assumption! The opposite is true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I guess the limit diverges to both negative and positive infinity.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Yes:) positive and negative infinity, because it alternates.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, Thanks. I guess my understand of factorials are a bit off.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3We 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
 one year ago
Best ResponseYou've already chosen the best response.0thx i learned something new just by watching lol

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3You would then agree that the numerator is greater, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do agree that the numerator is greater.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Good:) 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
 one year ago
Best ResponseYou've already chosen the best response.3(not necessarily n=1, but that series)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.3Anyway, good luck....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help. It really made since. I am now making forward progress on my assignments.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.3Good to hear that:)
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