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

In summary, the conversation discusses finding the limit of a sequence and explores various methods for solving the problem. The concept of factorials and their growth rate is also mentioned.
  • #1
Jimbo57
96
0

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?
 
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  • #2
Jimbo57 said:

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##.
 
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  • #3
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 [itex]n[/itex]s without a partner. Is there a different way to pair off terms so that the [itex]n[/itex]s have partners that "kill" them off?
 
  • #4
Zondrina said:
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.
 
  • #5
gopher_p said:
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 [itex]n[/itex]s without a partner. Is there a different way to pair off terms so that the [itex]n[/itex]s 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.
 
  • #6
Write the function as

[tex]4 \frac{n^2}{n(n-1)} \frac{2^{n-2}}{(n-2)!}[/tex]
 
  • #7
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, [itex]\forall[/itex] ε>0 [itex]\exists[/itex] N[itex]\in[/itex] 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
 
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  • #8
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
micromass said:
Write the function as

[tex]4 \frac{n^2}{n(n-1)} \frac{2^{n-2}}{(n-2)!}[/tex]

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?

[tex]4 \frac{n}{(n-1)} \frac{2^{n-2}}{(n-2)!}[/tex] This is what I get after cancelling n's. But as to show how this approaches 0, I'm totally lost.
 
  • #10
Jufro said:
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, [itex]\forall[/itex] ε>0 [itex]\exists[/itex] N[itex]\in[/itex] 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
This isn't for analysis Jufro, it's Calc 101 and I have yet to see any analysis.
 
  • #11
Zondrina said:
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.
 
  • #12
Jimbo57 said:
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?

[tex]4 \frac{n}{(n-1)} \frac{2^{n-2}}{(n-2)!}[/tex] 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?
 
  • #13
micromass said:
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.
 
  • #14
micromass said:
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.
 
  • #15
Just write out a few terms then you'll see.
 
  • #16
Jimbo57 said:
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".
 
  • #17
Zondrina said:
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.)
 
  • #18
DeIdeal said:
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!)!)!)...##
 
  • #19
You can use this little theorem that is very useful in situatons like this one.

Call your sequence [itex]a_n[/itex]

If [itex]a_n > 0[/itex]

you calculate this limit [itex]lim {a_{n+1} \over a_n}[/itex]

if that limit is beetween 0 and 1, but not 1,then [itex]a_n{\rightarrow} 0 [/itex]

if that limit is greater than 1 then [itex]a_n{\rightarrow} ∞ [/itex]

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



For example

[itex]n! \over e^n[/itex]

we calculate the limit [itex]{(n+1)! \over e^{n+1}}\over {n! \over e^n}[/itex]

and using simple properties of exponential and factorial we get

[itex]{n!(n+1) \over n!} { e^n \over e^n e} = {n+1 \over e} {\rightarrow}∞[/itex]

so [itex]{n! \over e^n}{\rightarrow}∞[/itex]

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

1. What is the purpose of finding the limit of a sequence?

The limit of a sequence is used to determine the behavior of the sequence as the number of terms increases. It helps to determine if the sequence converges or diverges, and if it converges, what the value of the limit is.

2. How do you find the limit of a sequence?

To find the limit of a sequence, you need to evaluate the expression for a large number of terms. This means plugging in larger and larger values of n and observing the resulting values. If the values approach a specific number as n increases, that number is the limit of the sequence.

3. What is the formula for finding the limit of a sequence?

The formula for finding the limit of a sequence is: lim n→∞ an, where 'a' is the general term of the sequence and n represents the number of terms in the sequence. This formula is used to evaluate the expression for a large number of terms to determine the limit.

4. What is the purpose of dividing by n! when finding the limit of a sequence?

Dividing by n! is used to simplify the expression and make it easier to evaluate for a larger number of terms. It also helps to determine the rate of growth of the sequence and whether it will converge or diverge.

5. Can you explain the process for finding the limit of n22n/(n!)?

To find the limit of n22n/(n!), we first need to observe the behavior of the sequence as n increases. We can simplify the expression to 2n/n!, and then plug in larger and larger values of n to see if the resulting values approach a specific number. If the values approach a specific number, that is the limit of the sequence. We can also use the formula lim n→∞ 2n/n! to evaluate the expression for a large number of terms and determine the limit.

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