Lightning question before the test : )

  • Thread starter frasifrasi
  • Start date
  • #1
frasifrasi
276
0
How can I show that the sequence { 4^(n)/n! }

converges to zero (by squeeze theorem)??

Thank you! : )
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
1,648
4
How can I show that the sequence { 4^(n)/n! }

converges to zero (by squeeze theorem)??

Thank you! : )

If the Squeeze Theorem is required as part of the proof, how about comparing

[tex]\frac{4^{n}}{n!}[/tex] < [tex]\frac{4^{n}}{n^{n}}[/tex] (for n > 1)

and showing that the latter goes to zero as n--> infinity?
 
  • #3
Gib Z
Homework Helper
3,352
7
Minor correction, but the "for n > 1" part isn't correct. You can find the correct value, or just say sufficiently large n.
 
  • #4
learningphysics
Homework Helper
4,099
6
as n->infinity n^n>n! so [tex]\frac{4^n}{n!} > \frac{4^{n}}{n^{n}}[/tex].

I think probably the idea being expressed was to use a constant in the denominator instead of n. [tex]\frac{4^n}{8^n}[/tex]. (or any constant^n in the denominator where the constant>4)

[tex]\frac{4^n}{n!}< \frac{4^n}{8^n}[/tex] for large enough n.
 
Last edited:
  • #5
dynamicsolo
Homework Helper
1,648
4
Minor correction, but the "for n > 1" part isn't correct. You can find the correct value, or just say sufficiently large n.

Whoops! Quite so -- I was only looking at the denominators. That needs to be n > 2 , doesn't it? (n=3 works...)
 

Suggested for: Lightning question before the test : )

Replies
2
Views
239
Replies
3
Views
304
  • Last Post
Replies
7
Views
545
Replies
7
Views
510
Replies
1
Views
534
Replies
2
Views
496
  • Last Post
Replies
7
Views
618
Replies
11
Views
437
  • Last Post
Replies
14
Views
699
  • Last Post
Replies
3
Views
770
Top