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

1. anonymous

I am trying to find the following limit. $\lim_{n \rightarrow \inf} \frac{ 2(n!) }{ (-6)^n }$

2. SolomonZelman

Factorial will grow much much bigger than any exponential growth (after some point).

3. anonymous

Is it because the factorial is multiplied by 2?

4. anonymous

I was under the assumption that the (-6)^n would grow faster than the factorial.

5. 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

6. anonymous

Then I guess the limit diverges to both negative and positive infinity.

7. SolomonZelman

Yes:) positive and negative infinity, because it alternates.

8. anonymous

Alright, Thanks. I guess my understand of factorials are a bit off.

9. SolomonZelman

We can play around and maybe apply Stirling's approximation: (Note, that the factorial is greater than its stirlings approximation) $$\large\color{black}{ \displaystyle n!\approx\frac{\sqrt{2\pi n}{~}n^n}{e^n} }$$ $$\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\frac{\sqrt{2\pi n}{~}n^n}{6^ne^n} }$$ $$\large\color{black}{ \displaystyle \frac{n!}{6^n}\approx\sqrt{2\pi n}\frac{{~}n^n}{(6e)^n} }$$

10. some.random.cool.kid

thx i learned something new just by watching lol

11. SolomonZelman

You would then agree that the numerator is greater, right?

12. anonymous

I do agree that the numerator is greater.

13. SolomonZelman

Good:) And I understand that you were applying the Diverghence test to $$\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{2(n!)}{(-6)^n}}$$

14. SolomonZelman

(not necessarily n=1, but that series)

15. anonymous

yep

16. 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: $$\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{(-6)^n}{n!}=e^{-6}= 1/e^6}$$

17. SolomonZelman

Anyway, good luck....

18. anonymous

Thanks for the help. It really made since. I am now making forward progress on my assignments.

19. SolomonZelman

Good to hear that:)

20. SolomonZelman

be well