# Try to plot it on mathematica for x=200

1. Jun 13, 2005

### steven187

hello all

how would one prove that $$\lim_{n\rightarrow\infty}\frac{x^{n+1}}{(n+1)!}=0$$ $$\forall x\in\Re$$ now when i try to plot it on mathematica for x=200, on the plot it displays that $$200^{n+1}>(n+1)!$$ how could it possibly converge for any value of x? the reason why i need to prove this is because im trying to show that a function is analytic, would somebody have an anology or a graphical explaination of what is meant by an analytic function?

Steven

Last edited: Jun 13, 2005
2. Jun 13, 2005

### shmoe

You could use Stirlings approximation to the factorial, but that's not necessary. Here's a brutal approach. First if 0<=x<=1 the result is obvious, so assume x>1. Fix x, let N be any integer larger than x^2. Then $$(N+M)!>(N)(N+1)\ldots(N+M)>N^M>(x^2)^M$$ for any positive integer M. Hence:

$$\frac{x^{N+M}}{(N+M)!}<\frac{x^{N+M}}{x^{2M}}=x^{N-M}$$

Thinking of n+1=N+M, your limit is now the limit as M->infinity and hence is zero.

If x is negative, the result will still hold, multiplying the terms by +/-1 the limit will still be zero.

3. Jun 13, 2005

### mathwonk

stirling is vast overkill. this obviously converges to 0, for any x. just notice that after n is larger than x, then each term is obtained by multiplying the previous one by something smaller than x/n, which is less than 1.

i.e. in each successive term the top is multiplied by the same old constant (x), but the bottom is multiplied by something getting larger and larger.

as soon as the bottom catches up to the top, the eventual future is clear.

Last edited: Jun 13, 2005
4. Jun 13, 2005

### shmoe

Overkill-yes, I only mentioned it because of some his other posts and he'll be meeting stirling's if he carries on studying zeta.

I also had meant to mention the attempt to graph this in mathematica when x=200. You probably weren't able to look very far. This will only start to turn around and decrease at n=199, where it has over 80 digits. I don't know mathematicas limits, but this is likely a problem (this is a good warning to beware any finite amount of data when dealing with an infinite limit). Try ploting log of this function if you want to look furthur.

5. Jun 13, 2005

### steven187

now what I want to conclude is that for large values of n,
$$n<n^2<e^n<n!<n^n$$
so would it be true to say that
$$\lim_{n\rightarrow\infty}\frac{n!}{n^n}=0$$
what i realise is that it all depends upon how fast each increases as n goes to infinity, the reason why i want to make such a conclusion is so that next time i come across a limite i will know weither the numerator is less than the denominator as n goes to infinity so that i will be able to know the limite of by heart, is there any special way of proving such limits, would L' hospital rule be the most general method to do such limits? well i will be reseaching into stirlings hopefully pretty soon is there any prior knowledge i need to know before i start on it?

Last edited: Jun 13, 2005
6. Jun 13, 2005

### TenaliRaman

Yes.

Thats perfectly correct and its a good way to make the analysis IMHO. However ofcourse not all limits can be gauged in this manner. There may be some limits where numerator and denominator, either proceed at the same rate or we dont see an immediate way to compare their rates. (simple example being those functions which are not monotonically increasing or decreasing)

Not such a good idea. Its better to do analysis each time you find a limit. Nonetheless, even if you dont byheart, by enough practice you can evaluate limits in under a minute (given its moderately hard).

Special way to do all limits? nope. You can use taylor series, and you can use your idea of rate of increase and some more stuff.

L'Hospital is the cut down version of taylor series method, so you can easily say its not applicable where you cannot use taylor series.

Stirling's formula is an approximation for factorial and nothing else. However, if u want to understand the proof of it then for the method i recall, you need
simple algebra,
integration and
taylor series for log
Along with it you need to use the (not-so) famous wallis formula. The proof of which again is simple. Both of these proofs are available on net, a simple google search should lead you to them.

-- AI

7. Jun 14, 2005

### steven187

im trying to draw a picture in my head so i could refer to it when needed

thanxs