Convergence of n!/n^n Sequence

  • Thread starter ehrenfest
  • Start date
In summary, to prove that the sequence \frac{n!}{n^n} converges to 0, we can split off some of the earlier terms in the product and show that they tend to zero. This can be done by choosing N such that 2^-N is smaller than the remaining terms in the product.
  • #1
ehrenfest
2,020
1
[SOLVED] proving a seqeunce converges

Homework Statement


Prove that the sequence [itex]\frac{n!}{n^n}[/itex] converges to 0.

Homework Equations


The Attempt at a Solution


Given [itex]\epsilon > 0[/itex], how do I find N? We know that the nth term is less than or equal to
[tex]\left(\frac{n-1}{n}\right)^{n-1}[/tex]
but that really does not help.
 
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  • #2
… in a fight, always pick off the smallest ones …

Hi ehrenfest! :smile:

Hint: write it (1/n)(2/n)(3/n)…(n/n);

the early bits go down much faster than the later bits; so can you see a way of splitting off some of the early bits, and prove that they tend to zero? :smile:
 
  • #3
tiny-tim said:
Hi ehrenfest! :smile:

Hint: write it (1/n)(2/n)(3/n)…(n/n);

the early bits go down much faster than the later bits; so can you see a way of splitting off some of the early bits, and prove that they tend to zero? :smile:

Yes, I was trying to do something like that. If I can show that any factor in that product goes to 0, then the whole thing has to because the rest will be less than 1. If I collect the first n/2 ceiling terms, then I get (1/2)^(n/2) if n is even, and that needs to got to zero, doesn't it? I see, thanks. When n is odd we get something similar.
 
  • #4
That's right! :smile:

(And don't forget, your proof should begin something like "For any epsilon, choose N such that 2^-N …")
 

1. What is the definition of a convergent sequence?

A convergent sequence is a sequence of numbers in which the terms become increasingly closer to a single number, known as the limit. In other words, as the sequence progresses, the terms get closer and closer to the limit, and eventually, the terms will be infinitely close to the limit.

2. How do you prove that a sequence converges?

To prove that a sequence converges, you must show that the terms of the sequence get closer and closer to the limit as the sequence progresses. This can be done by showing that the difference between the terms and the limit approaches zero as the sequence progresses.

3. What is the role of the epsilon-N definition in proving convergence?

The epsilon-N definition, also known as the limit definition of convergence, is a mathematical statement that defines the concept of convergence. It states that for a sequence to converge to a limit L, the terms of the sequence must eventually get arbitrarily close to L, meaning that for any positive number ε (epsilon), there exists a term in the sequence after which all the terms are within ε distance from the limit L.

4. Can a sequence have multiple limits?

No, a sequence can only have one limit. If a sequence has more than one limit, then it is said to be divergent, meaning that it does not converge to a single number. However, a sequence can approach a limit from different directions, such as from above or below, which may give the appearance of multiple limits.

5. Are all convergent sequences also bounded?

Yes, all convergent sequences are bounded. This means that there is a fixed value that the terms of the sequence cannot exceed. This fixed value is known as the bound, and it is usually the limit of the sequence. A sequence that is unbounded, meaning that the terms have no limit or bound, is not convergent.

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