Proving $\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0$ - Tutoring Maths Resits

  • Thread starter Zurtex
  • Start date
  • Tags
    Limit
In summary: N}which is fine, but it asks to him to do it by proving by definition that: \lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0And he remembers how to do it, so he just needs to finish the proof. In summary, George needs to prove that: n^{n-1} \geq n! \quad \forall n > 1
  • #1
Zurtex
Science Advisor
Homework Helper
1,120
1
I'm doing some tutoring tomorrow morning for a girl who needs to do some maths resits. I was looking through the papers and you need to prove that:

[tex]n^{n-1} \geq n! \quad \forall n > 1 \; \text{and} \; n \in \mathbb{N}[/tex]

Which is fine, but it asks to you to do it by proving by definition that:

[tex]\lim_{n \rightarrow \infty} \frac{n!}{n^n} = 0[/tex]

And I have to admit that I can't quite remember how to do this one, if someone could point me in the right direction that would be great, I remember seeing a proof for it so I'm sure it'll come back to me.
 
Physics news on Phys.org
  • #2
Split it into a part that obviously goes to zero, and a part that is obviously bounded!
 
  • #4
iNCREDiBLE said:
Stirling's formula/approximation works fine too.
Too far ahead for the material she is expected to know and my brain isn't working at the moment, I don't know what you mean Hurkyl
 
  • #5
I doubt this would pass as a proof, but here 1/n goes to zero and the second limit has an equal number of factors in the nominator as denominator. The highest factors are equal (n) but in the nominator, they decrease while they don't in the denominator.

[tex]\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n^n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n \cdot n^{n - 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{n^{n - 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\mathop {\lim }\limits_{n \to \infty } \frac{{\overbrace {n \cdot \left( {n - 1} \right) \cdot \ldots \cdot 2}^{n - 1}}}{{\underbrace {n \cdot n \cdot \ldots \cdot n}_{n - 1}}}=0[/tex]
 
  • #6
[tex]
\left( \frac{n \cdot (n-1) ... 2} {n \cdot n ... n} \right) \frac{1} {n}
[/tex]

The stuff in the big brackets is less greater than zero and less than 1 for n > 2.

Regards,
George
 

Q1: What is the meaning of the limit in this statement?

In this statement, the limit refers to the behavior of the expression $\frac{n!}{n^n}$ as the value of $n$ approaches infinity.

Q2: How is it possible to prove this limit using mathematical reasoning?

This limit can be proven using the definition of a limit, which states that for any $\epsilon > 0$, there exists a positive integer $N$ such that for all $n > N$, the expression $\left|\frac{n!}{n^n} - 0\right| < \epsilon$ is satisfied. In other words, as $n$ gets larger and larger, the value of $\frac{n!}{n^n}$ gets closer and closer to 0.

Q3: What is the significance of this limit in mathematics?

This limit is significant because it shows that the factorial function grows at a much slower rate than the exponential function. In other words, as $n$ approaches infinity, the value of $\frac{n!}{n^n}$ approaches 0, indicating that the factorial function becomes relatively insignificant compared to the exponential function.

Q4: Can this limit be proven using various mathematical techniques?

Yes, this limit can be proven using various techniques, such as the squeeze theorem, which states that if two functions, $f(n)$ and $g(n)$, both approach the same limit $L$ as $n$ approaches a specific value, and another function $h(n)$ always falls between them, then $h(n)$ also approaches $L$ as $n$ approaches that value. In this case, we can use the fact that $0 \leq \frac{n!}{n^n} \leq \frac{1}{n}$ for all $n > 0$, and since $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$, the squeeze theorem can be applied to prove the desired limit.

Q5: How can this limit be applied in real-world scenarios?

This limit has applications in various fields of science and engineering, such as in the analysis of algorithms, where it can be used to determine the efficiency of a particular algorithm. It can also be used in statistics and probability to calculate the probability of a particular event occurring in a large number of trials. Additionally, this limit is essential in understanding the concept of asymptotic behavior, which is used to describe how functions behave as their input values approach infinity.

Similar threads

MHB Exploring $\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$
    • Calculus
Replies
2
Views
947
MHB Why Does $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$?
    • Calculus
Replies
6
Views
1K
Show that the limit (1+z/n)^n=e^z holds
    • Calculus and Beyond Homework Help
Replies
6
Views
472
MHB Limits of Recurrence Relations with $0<b<a$
    • Calculus
Replies
2
Views
919
MHB Evaluate Limit: $\frac{0}{0}$ - Can Someone Help?
    • Calculus
Replies
4
Views
1K
MHB Proof of $\lim_{{n}\to{\infty}}a_n=\lim_{{n}\to{\infty}}b_n$ with $\epsilon-N$
    • Calculus
Replies
7
Views
2K
I Gradient With Respect to a Set of Coordinates
    • Calculus
Replies
3
Views
1K
I Prove that the tail of this distribution goes to zero
    • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
MHB Why is $U_n$ greater than $\frac{A}{n}$ for large enough values of $n$?
    • Calculus
Replies
4
Views
1K
MHB Calculate Upper Bound for $\displaystyle a_{n}$ in Binomial Limit Evaluation
    • Calculus
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top