Find a Sequence to Make lim(An/An+1)=∞

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Discussion Overview

The discussion revolves around the possibility of finding a sequence \( A_n \) such that the limit \( \lim_{n \to \infty} \frac{A_n}{A_{n+1}} = \infty \) while also satisfying \( \lim_{n \to \infty} A_n = \infty \). Participants explore various sequences and mathematical expressions to address this question.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant initially seeks help in proving whether such a sequence exists.
  • Another participant suggests analyzing \( \frac{A_n}{A_n + 1} \) but clarifies that the focus is on \( \frac{A_{n+1}}{A_n} \).
  • A proposed approach is to set \( \frac{A_{n+1}}{A_n} = n \), leading to the recursive relation \( A_{n+1} = n A_n \). Participants wonder if a solution can be found for this relation.
  • It is suggested that the limit could be evaluated as \( e \cdot \lim(n+1) \) resulting in \( e \cdot \infty = \infty \), contingent on the sequence being of the form \( C n^n \) for some constant \( C \neq 0 \).
  • Another example provided is \( A_n = C n! \), which leads to \( \frac{A_{n+1}}{A_n} = n + 1 \).
  • A participant corrects their earlier example, confirming that \( A_n = C n! \) is the intended sequence while also acknowledging that \( A_n = C n^n \) could work as well.

Areas of Agreement / Disagreement

Participants explore multiple approaches and examples, but there is no consensus on a definitive sequence that satisfies the original condition. Various models and expressions are proposed, indicating a lack of resolution.

Contextual Notes

Participants express uncertainty regarding the existence of a sequence that meets the specified limit conditions and the implications of their proposed solutions. The discussion includes different mathematical forms and their evaluations without settling on a singular approach.

esuahcdss12
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Hey

suppose I have sequence An

limAn,n→∞ = ∞

Is it possible to find a sequence which makes:

lim (An/An+1) ,n →∞ = ∞?

I tried to search a sequence like that and could not find, but I don't know how to prove that this is
can not be happening.
could you help please?
 
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I'm assuming you mean to study $\frac{A_n}{A_n + 1}$. Note that $\lim_{n \to \infty}\frac{A_n}{A_n + 1} = \lim_{n \to \infty}\frac{1}{1 + \frac{1}{A_n}}$.
 
Krylov said:
I'm assuming you mean to study $\frac{A_n}{A_n + 1}$. Note that $\lim_{n \to \infty}\frac{A_n}{A_n + 1} = \lim_{n \to \infty}\frac{1}{1 + \frac{1}{A_n}}$.

no, I meant (A(n+1)/A(n))
 
Hi esuahcdss12!

So we want $\frac{A_{n+1}}{A_n}$ to diverge.
How about setting it for instance to be equal to $n$, which is the simplest expression that diverges.
That means we get $\frac{A_{n+1}}{A_n}=n\quad\Rightarrow\quad A_{n+1} = n A_n$.
Can we find a solution for that? (Wondering)
 
I like Serena said:
Hi esuahcdss12!

So we want $\frac{A_{n+1}}{A_n}$ to diverge.
How about setting it for instance to be equal to $n$, which is the simplest expression that diverges.
That means we get $\frac{A_{n+1}}{A_n}=n\quad\Rightarrow\quad A_{n+1} = n A_n$.
Can we find a solution for that? (Wondering)

It could be done only if $$n^n$$ and then after some calculation of $$\frac{A(n+1)}{An}$$
we get that the limit is e*lim(n+1) which is e *∞ = ∞
Is that correct?
 
esuahcdss12 said:
It could be done only if $$n^n$$ and then after some calculation of $$\frac{A(n+1)}{An}$$
we get that the limit is e*lim(n+1) which is e *∞ = ∞
Is that correct?

Let's make that 'only if' with $Cn^n$ for some constant $C\ne 0$. Otherwise it's just an example (which it is anyways).
And yes, that is how the limit would be evaluated. (Nod)
 
Another example would be:

$$A_{n}=Cn!$$

So that:

$$\frac{A_{n+1}}{A_{n}}=\frac{C(n+1)!}{Cn!}=n+1$$
 
MarkFL said:
Another example would be:

$$A_{n}=Cn!$$

Oh wait! (Wait)
I think my example should be $A_n=Cn!$ instead of $A_n=Cn^n$. (Blush)
Then again $A_n=Cn^n$ also works.
 

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