Convergence of Factorial Function: Is Stirling Approximation Necessary?

In summary: I'm sorry, I meant that the terms in the geometric sequence should be greater than or equal to the corresponding terms in the original sequence.As for your second question, you need to show that all of the terms are greater than or equal to the corresponding terms in the original sequence. This will guarantee that the geometric sequence is an upper bound for the original sequence.
  • #1
Seydlitz
263
4

Homework Statement


$$\lim_{n \to \infty}\frac{10^n}{n!}$$

Homework Equations


Do I need to use something as advanced as Stirling approximation?
This question appears in page 5 of Boas Mathematical Methods, there's no prior discussion on factorial or convergence test and etc.

The Attempt at a Solution


I know the limit should be 0 because factorial got larger than power as n gets larger, but I'm not convinced with this treatment. What if it's not 10 but some arbitrary number. I've tried to write the terms and to cancel the factor but the sequence is still mainly irregular.

$$\frac{10}{1},\frac{50}{1},\frac{500}{3},\frac{1250}{3},\frac{2500}{3},\frac{12500}{9},...$$

Thank You
 
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  • #2
No. If you look at the sequence ##a_n = \frac{10^n}{n!}##, you can see that ##a_n = a_{n-1}\frac{10}{n}##. It should be clear that for n>10, the sequence decreases monotonically. It's also bounded below by 0, so you know it has to converge.

To prove that it converges to 0, try using the sandwich theorem. You should be able to find a geometric sequence that bounds it from above and converges to 0.
 
  • #3
vela said:
No. If you look at the sequence ##a_n = \frac{10^n}{n!}##, you can see that ##a_n = a_{n-1}\frac{10}{n}##. It should be clear that for n>10, the sequence decreases monotonically. It's also bounded below by 0, so you know it has to converge.

To prove that it converges to 0, try using the sandwich theorem. You should be able to find a geometric sequence that bounds it from above and converges to 0.

Ok I understand your explanation but I don't know yet how to find the upper-bound sequence. The explanation for this problem set doesn't mention on how to know the convergence of a sequence. It merely says find the limit. Only in the next couple of pages can you find explanation on convergences and divergence.
 
  • #4
Hint: Try showing that ##a_{12} < \left(\frac{10}{11}\right)^2 a_{10}##.
 
  • #5
So it is possible to reduce that inequality into ##\frac{10^{12}}{12!} < \frac{10^{12}}{11(11)!}##. Furthermore this can be reduced to ##\frac{10^{12}}{12} < \frac{10^{12}}{11}##. Because the denominator is higher the term on the left is smaller.

The fact can be generalized to ##\frac{10^n}{n!}<\frac{10^n}{(n-1)(n-1)!}## Is the sequence on the right the upper bound you're referring to? Can we take ##\frac{10^n}{(n-1)!}## also as an upper bound?

What can we learn from the process? If the individual term of one sequence is greater than the other and if it converges will it be automatically its upper bound also?
 
  • #6
Seydlitz said:
So it is possible to reduce that inequality into ##\frac{10^{12}}{12!} < \frac{10^{12}}{11(11)!}##. Furthermore this can be reduced to ##\frac{10^{12}}{12} < \frac{10^{12}}{11}##. Because the denominator is higher the term on the left is smaller.

The fact can be generalized to ##\frac{10^n}{n!}<\frac{10^n}{(n-1)(n-1)!}## Is the sequence on the right the upper bound you're referring to? Can we take ##\frac{10^n}{(n-1)!}## also as an upper bound?
What would be the point? The new sequence is simply 10 times the original sequence. If you could show directly that one sequence converged, you could use the same method for the other.

I suggested that you compare the original sequence to an appropriate geometric sequence. Do you know what a geometric sequence is? If not, you should look it up.

What can we learn from the process? If the individual term of one sequence is greater than the other and if it converges will it be automatically its upper bound also?
Given that you're using the Boas text, I assume that you've already taken a year of calculus. This material on sequences should be review for you. If you don't remember it well, you should go back and look over it again.
 
  • #7
vela said:
What would be the point? The new sequence is simply 10 times the original sequence. If you could show directly that one sequence converged, you could use the same method for the other.

I suggested that you compare the original sequence to an appropriate geometric sequence. Do you know what a geometric sequence is? If not, you should look it up.

I know what a geometric sequence is but I don't know you could actually proceed in such a way and I just realized now that want me to find geometric sequence.

Will the geometric sequence ##(\frac{10}{11})^{2n}## be acceptable? We can take ##\frac{10^{10}}{10!}## as it's first term. Clearly this geometric sequence converges to 0 as with every geometric sequence, and the first term is the maximum value the original sequence can get? So another bound.
 
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  • #8
Seydlitz said:

Homework Statement


$$\lim_{n \to \infty}\frac{10^n}{n!}$$


Homework Equations


Do I need to use something as advanced as Stirling approximation?
This question appears in page 5 of Boas Mathematical Methods, there's no prior discussion on factorial or convergence test and etc.

The Attempt at a Solution


I know the limit should be 0 because factorial got larger than power as n gets larger, but I'm not convinced with this treatment. What if it's not 10 but some arbitrary number. I've tried to write the terms and to cancel the factor but the sequence is still mainly irregular.

$$\frac{10}{1},\frac{50}{1},\frac{500}{3},\frac{1250}{3},\frac{2500}{3},\frac{12500}{9},...$$

Thank You
The Stirling Asymptotic Relation isn't that advanced. You could do $$\lim_{n\rightarrow +\infty}\frac{10^n}{n!}=\lim_{n\rightarrow +\infty}\frac{10^n}{\sqrt{2\pi n}(\frac{n}{e})^n}.$$

However, let's think about this logically. Which one grows significantly faster than the other?

##10^{70}## looks big, right? ##70!## is larger than googol. Factorials grow MUCH faster.
 
  • #9
Seydlitz said:
Will the geometric sequence ##(\frac{10}{11})^{2n}## be acceptable? We can take ##\frac{10^{10}}{10!}## as it's first term. Clearly this geometric sequence converges to 0 as with every geometric sequence, and the first term is the maximum value the original sequence can get? So another bound.
Can you prove the terms are greater than the terms in the original sequence? If so, it's acceptable.
 
  • #10
vela said:
Can you prove the terms are greater than the terms in the original sequence? If so, it's acceptable.

Do I need to show that all of the terms are greater or equal than the original or will showing the first term is equal with the maximum value of the original sequence suffice?
 
  • #11
You have to show all terms after some point are bounded above by the geometric sequence.
 

FAQ: Convergence of Factorial Function: Is Stirling Approximation Necessary?

What is the factorial function?

The factorial function is a mathematical function that is used to calculate the product of all positive integers less than or equal to a given number. It is denoted by the symbol "!" and is read as "n factorial". For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

What is the limit of the factorial function?

The limit of the factorial function, denoted by "n!" or "n→∞", is the largest number that can be calculated using the factorial function. As n increases, the value of n! also increases exponentially, approaching infinity as n approaches infinity.

What is the value of 0 factorial?

The value of 0 factorial, denoted by 0!, is defined to be equal to 1. This definition is based on mathematical conventions and is necessary for certain mathematical formulas and equations to work properly.

How is the factorial function used in mathematics?

The factorial function is used in various mathematical calculations, such as combinations and permutations, probability calculations, and in the Taylor series expansion of certain functions. It is also used in the study of number theory and in various fields of science and engineering.

What is the relationship between the factorial function and the gamma function?

The gamma function, denoted by Γ(n), is an extension of the factorial function to real and complex numbers. It is closely related to the factorial function, with Γ(n+1) = n!. The gamma function is used in various areas of mathematics, such as complex analysis and probability theory.

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