Convergence of Factorial Function: Is Stirling Approximation Necessary?

[Seydlitz]

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

 

[vela]

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.

 

[Seydlitz]

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.

 

[vela]

Hint: Try showing that ##a_{12} < \left(\frac{10}{11}\right)^2 a_{10}##.

 

[Seydlitz]

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?

 

[vela]

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.

 

[Seydlitz]

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.

 

[Mandelbroth]

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.

 

[vela]

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.

 

[Seydlitz]

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?

 

[vela]

You have to show all terms after some point are bounded above by the geometric sequence.