Find Limit of Sequence: n^2*2^n/n!

[khari]

Problem: Find the limit of the sequence

$$a_{n}=\frac{n^{2}2^{n}}{n!}$$

My first thought was to say that

$$0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}$$

and by squeeze theorem, since $$\frac{x^n}{n!}=0$$ for all real x, my original limit must be 0 as well. Now all I need to do is prove that $$n^{2}2^{n} \leq x^n$$, which is where I'm stuck. Can anyone give me a hand? Thanks.

 

[EnumaElish]

Rearrange to n^2 < (x/2)^n.

 

[dynamicsolo]

Problem: Find the limit of the sequence

$$a_{n}=\frac{n^{2}2^{n}}{n!}$$

My first thought was to say that

$$0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}$$

and by squeeze theorem, since $$\frac{x^n}{n!}=0$$ for all real x, my original limit must be 0 as well. Now all I need to do is prove that $$n^{2}2^{n} \leq x^n$$, which is where I'm stuck. Can anyone give me a hand? Thanks.

What is x? Is $$0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}$$ even true?

You might break up the numerator and denominator into their respective factors and look at the ratios that are formed. I believe you'll find that some have specific values, others have finite values in the limit n-> inf. , and others go to zero in the limit.

 

[EnumaElish]

What is x? Is $$0\leq \frac{n^{2}2^{n}}{n!} \leq \frac{x^n}{n!}$$ even true?

Not for any x and any n; but I think n^2 2^n < x^n as n --> +infinity.

 

[khari]

Rearrange to n^2 < (x/2)^n.

Sorry if it seems stupid, but I'm horrible at proofs, and am still coming up a bit short on proving $$n^2 \leq {(\frac{x}{2})}^n$$

I mean it seems fairly obvious that this is true to me, and more obvious still when I begin writing it out term by term for increasing n, but is this good enough to state it conclusively?

 

[Dick]

dynamicsolo's advice is a bit better. Write it as n*n*4*2^(n-2)/(n*(n-1)*(n-2)!). Does that suggest anything?

 

[khari]

dynamicsolo's advice is a bit better. Write it as n*n*4*2^(n-2)/(n*(n-1)*(n-2)!). Does that suggest anything?

I must be missing something obvious, because I've tried that as well with pretty much no success.

 

[dynamicsolo]

I must be missing something obvious, because I've tried that as well with pretty much no success.

I'm suggesting that you write each factor in the numerator over a factor in the denominator and then consider each ratio you've formed. An example:

(e^n)/(n!) = (e·e·...·e)/[n·(n-1)·...·3·2·1] ;

there are n factors each in the numerator and denominator, so you can also write this as

(e/n)·(e/n-1)·...·(e/3)·(e/2)·(e/1).

In the case of our ratio, there are (n+2) factors in the numerator and only n in the denominator, so we will need to set aside two factors upstairs -- two of the 2s may be best. Now look at your ratio as a product of factors and recall that the limit of a product of terms is the product of the limits of the terms.

Remember also that since we want the limit of the sequence a-sub-n, we are looking for the limit of this product as n-> infinity.

 

[Dick]

What are the limits of n/n, n/(n-1) and coup de grace 2^(n-2)/(n-2)!, I thought you said you knew x^n/n! -> 0. x=2 and if n->infinity, n-2->infinity.

 

[khari]

I got it. Thank you all, you've been a big help.

 

[dynamicsolo]

*thuds head with heel of hand, causing skull to emit dull, hollow sound*

I just thought of something else: what is

lim n->inf. of abs($$\frac{a_{n+1}}{a_{n}}$$)?

We know the terms of this sequence are positive and you'll find this limit tends to zero. Thus the limit of the sequence is zero.