Proving \lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0

In summary, we are trying to prove that the limit as n approaches infinity of p(n)/e^n is equal to 0 for a polynomial p(x) with real coefficients. The attempt at a solution suggests using series and L'Hospital's rule, but since it has not been proven in class, another approach is needed. The idea is to show that a_kn^k/e^n goes to 0 as n approaches infinity, and since this is the largest term in the polynomial, it can be used to find a suitable N for all the terms. However, it may be necessary to take more care for negative a_k's.
  • #1
Zhalfirin88
137
0

Homework Statement



Prove [itex] \lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0[/itex] where [itex] p(x) = a_k x^k + ... + a_1 x + a_0 [/itex] (with real coefficients [itex] a_i [/itex] in [itex] \mathbb{R} ) [/itex]


The Attempt at a Solution



I thought about using series to try and prove this, but I couldn't get it to work out and I think there is an easier way.

[itex] \frac{p(n)}{e^n} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k n^k}{\sum_{n=0}^\infty \frac{n^k}{k!}} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k}{\sum_{n=0}^\infty \frac{1}{k!}} [/itex]
 
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  • #2
Zhalfirin88 said:

Homework Statement



Prove [itex] \lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0[/itex] where [itex] p(x) = a_k x^k + ... + a_1 x + a_0 [/itex] (with real coefficients [itex] a_i [/itex] in [itex] \mathbb{R} ) [/itex]


The Attempt at a Solution



I thought about using series to try and prove this, but I couldn't get it to work out and I think there is an easier way.

[itex] \frac{p(n)}{e^n} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k n^k}{\sum_{n=0}^\infty \frac{n^k}{k!}} [/itex] = [itex] \frac{ \sum_{n=0}^\infty a_k}{\sum_{n=0}^\infty \frac{1}{k!}} [/itex]

Replace n by a continuous variable x and use L'Hospital's rule.
 
  • #3
I don't think I'm allowed to use L'Hospital's rule, we have not proven it in class yet.
 
  • #4
Could you show that [itex]a_kn^k/e^n[/itex] goes to 0 as n goes to infinity? Then each lower order term would go to 0, as well. What tests/methods are you allowed to use? Does this have to be strict [itex]\epsilon - \delta[/itex] from the defimition of limit?
 
  • #5
No we are proving things based on the definition of convergence, for instance, our proofs start off as "Let [itex] \epsilon > 0 [/itex] be arbitrary, and choose N > ... such that for all n > N ... etc. We haven't done epsilon delta proofs yet.
 
  • #6
Whoops! yes, [itex] \epsilon - N[/itex] is what I meant. But anyway, could you show that [itex]a_kn^k/e^n[/itex] goes to 0? For instance, could you do something like this:
Let [itex]a = sup \{|a_i| : 0 \leq i \leq k \}[/itex] then [itex] a_kn^k/e^n \leq an^k/e^n [/itex] and then show that [itex] an^k/e^n [/itex] goes to 0? Then deduce that the lower order terms go to 0? Of course, you are going to have to take more care than I did because 1)you'll need to find an N and 2)you'll need to take care of the times when you have negative [itex]a_k[/itex]'s. Anyway, am I making sense?
 
  • #7
Actually, since [itex] a_kn^k/e^n \leq an^k/e^n [/itex] is the largest term of the polynomial, couldn't you just find a N that suits this expression? Because [itex] p(n)/e^n [/itex] can be written as [itex] \frac{a_0}{e^n} + \frac{a_1 n}{e^n} + ... + \frac{a_k n^k}{e^n} [/itex], and since you choose n > N always, then the N that you pick for the largest term will be true for all terms, which will satisfy the definition, correct?
 
  • #8
Correct; that's what I had in mind. Now, find that N. Is this an Analysis class? I had terrible trouble proving things like this because it is so freaking hard to know what you are allowed to use. For example, L'Hospital's is off the table, but what can you use? The obvious answer, at first, is that you can only use stuff that you have already learned in the analysis class but I'd be willing to bet you haven't defined e^x, yet, is this correct? Do you find this a confusing as I did?
 
  • #9
Yes, an analysis class, but yeah it's a little awkward, especially since the professor used a derivative to prove some lemma or something, then goes and says "we shouldn't be using derivatives yet, but I'll use it anyways" :) And no we have not defined e^x yet, correct
 

Related to Proving \lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0

1. What does the equation "lim_{n\rightarrow\infty} \frac{p(n)}{e^n} = 0" mean?

The equation represents the limit of the ratio between a function p(n) and the exponential function e^n as n approaches infinity. This means that as n gets larger and larger, the value of p(n) relative to e^n approaches 0.

2. How is this limit proven?

There are several different methods for proving this limit, including using the squeeze theorem, the definition of a limit, and L'Hospital's rule. The specific method used will depend on the form of the function p(n).

3. What does it mean for a limit to equal 0?

When a limit equals 0, it means that the function being evaluated is approaching 0 as the input (in this case, n) becomes very large. This does not necessarily mean that the function itself is equal to 0, but rather that it is becoming increasingly smaller.

4. Can this limit ever be equal to a value other than 0?

Yes, depending on the function p(n) and the value of n, the limit may be equal to a value other than 0. However, for this specific equation, the limit will always approach 0 as n gets larger.

5. What is the significance of proving this limit?

Proving this limit is important in understanding the behavior of functions as n approaches infinity. It also has applications in various fields of mathematics and science, such as in the study of exponential growth and decay.

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