# Limit of a Sequence

1. Jul 31, 2013

### Jimbo57

1. The problem statement, all variables and given/known data
Find the limit of n22n/(n!)

2. Relevant equations

3. The attempt at a solution

First I expand out 2n/(n!) = (2/1)(2/2)(2/3)(2/4)...(2/n) which gets increasingly small as n increases. Now, where does the n2 fit into this? I know the limit to be 0 but I can't get passed this point. Any pointers?

2. Jul 31, 2013

### Zondrina

For sufficiently large n (try n=20 which is quite large actually) : $n! > n^2 2^n$.

3. Jul 31, 2013

### gopher_p

It looks like you are on to the right idea of writing the numerator and denominator as products of several terms and then pairing off terms. However your the way that you've paired off the terms leaves a couple of $n$s without a partner. Is there a different way to pair off terms so that the $n$s have partners that "kill" them off?

4. Jul 31, 2013

### Jimbo57

Is this generally enough to show the limit of a sequence approaches 0? I put it as my answer anyways since it seems pretty well known.

5. Jul 31, 2013

### Jimbo57

Hmmm are you talking about the n2 ? I don't know how else to expand the 2n/n! to cancel out n's unfortunately.

6. Jul 31, 2013

### micromass

Staff Emeritus
Write the function as

$$4 \frac{n^2}{n(n-1)} \frac{2^{n-2}}{(n-2)!}$$

7. Jul 31, 2013

### Jufro

If this is for some analysis course then you have to go back to the definition.

For a sequence an to have a limit L, $\forall$ ε>0 $\exists$ N$\in$ Z s.t. n>N then d(an, L) < ε.

You can start by assuming that a limit L exists, if you find one then the limit does exist and it is unique. If you don't find one, then you have a contradiction in the statement that you have a limit L.

So with that assumption it is easy to say the the limit of a positive sequence must be non-negative. From that point you can try to go through the cases. L=0 or strictly L>0.

L=0 is a good place to start, leaving you with d(n22n/n!, 0)<ε. Can you find a N which would make this true?

It may be easier at this point to switch from the general d notation to euclidean distance and end up with the equation N22N/N! = ε

8. Jul 31, 2013

### Zondrina

Sorry to ask, but maybe I don't know 100%. I believe factorials are the fastest growing elementary function ( unless you do multiple factorials such as n!!! )?

9. Jul 31, 2013

### Jimbo57

I'm sort of confused where the 4 came from. Is that just the result of pairing up the first 2, then showing the third 2 term as 2n-2?

$$4 \frac{n}{(n-1)} \frac{2^{n-2}}{(n-2)!}$$ This is what I get after cancelling n's. But as to show how this approaches 0, I'm totally lost.

10. Jul 31, 2013

### Jimbo57

This isn't for analysis Jufro, it's Calc 101 and I have yet to see any analysis.

11. Jul 31, 2013

### micromass

Staff Emeritus
Depends on what you mean with elementary, but $n^n$ grows faster.

12. Jul 31, 2013

### micromass

Staff Emeritus
What does $\frac{n}{n-1}$ converge to?

13. Jul 31, 2013

### Zondrina

Log, exponential, polynomial, trigonometric and their inverses, and factorials.

Thanks though, indeed $n^n$ grows faster.

14. Jul 31, 2013

### Jimbo57

$\frac{n}{n-1}$ converges to 1

Now, dealing with the factorial is where I get lost.

$\frac{2^{n-2}}{(n-2)!}$ converges to 0, but I'm not sure how.

15. Aug 1, 2013

### dirk_mec1

Just write out a few terms then you'll see.

16. Aug 1, 2013

### HallsofIvy

Staff Emeritus
In Europe, what would be called "Calculus" in the United States" is typically called "Analysis".

17. Aug 1, 2013

### DeIdeal

Just a quick sidenote, but n!! and n!!! are usually taken to mean double and triple factorials instead of (n!)! or something, that is, the product of every second or every third integer (for example n!!=n(n-2)(n-4)...4*2 for even n) as opposed to all integers, so n! "grows faster" than n!! or n!!!.

(Writing so many consecutive factorials looks really rude :[ I don't want to seem like I'm yelling at everyone.)

18. Aug 1, 2013

### Zondrina

Yes this is true, I was intending to mean $(((n!)!)!)...$ which grows much faster than $n^n$.

Though I was fiddling around on wolfram for fun and in general $n^{n^{n^{.^{.}}}} > (((n!)!)!)...$

19. Aug 1, 2013

### Dansuer

You can use this little theorem that is very useful in situatons like this one.

Call your sequence $a_n$

If $a_n > 0$

you calculate this limit $lim {a_{n+1} \over a_n}$

if that limit is beetween 0 and 1, but not 1,then $a_n{\rightarrow} 0$

if that limit is greater than 1 then $a_n{\rightarrow} ∞$

if that limit is 1 you can't say anything.

For example

$n! \over e^n$

we calculate the limit ${(n+1)! \over e^{n+1}}\over {n! \over e^n}$

and using simple properties of exponential and factorial we get

${n!(n+1) \over n!} { e^n \over e^n e} = {n+1 \over e} {\rightarrow}∞$

so ${n! \over e^n}{\rightarrow}∞$

It's very useful when you have two infinities one over the other. It works wounderfully when those infinities are exponentials or factorials.