Limit of Sequence Homework: Find n22n/(n!)

[Jimbo57]

Homework Statement

Find the limit of n²2ⁿ/(n!)

Homework Equations

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?

 

[STEMucator]

Homework Statement

Find the limit of n²2ⁿ/(n!)

Homework Equations

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?

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

 

[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 ns without a partner. Is there a different way to pair off terms so that the ns have partners that "kill" them off?

 

[Jimbo57]

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

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.

 

[Jimbo57]

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 ns without a partner. Is there a different way to pair off terms so that the ns have partners that "kill" them off?

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

 

[micromass]

Write the function as

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

 

[Jufro]

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

For a sequence a_n to have a limit L, \forall ε>0 \exists N\in Z s.t. n>N then d(a_n, 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 N²2^N/N! = ε

 

[STEMucator]

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! )?

 

[Jimbo57]

Write the function as

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

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 2^n-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.

 

[Jimbo57]

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

For a sequence a_n to have a limit L, \forall ε>0 \exists N\in Z s.t. n>N then d(a_n, 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 N²2^N/N! = ε

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

 

[micromass]

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! )?

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

 

[micromass]

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 2^n-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.

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

 

[STEMucator]

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

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

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

 

[Jimbo57]

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

$\frac{n}{n-1}$ 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.

 

[dirk_mec1]

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

 

[HallsofIvy]

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

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

 

[DeIdeal]

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! )?

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.)

 

[STEMucator]

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.)

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 $n^{n^{n^{.^{.}}}} > (((n!)!)!)...$

 

[Dansuer]

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

Call your sequence a_n

If a_n &gt; 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.