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

  • Thread starter Thread starter esuahcdss12
  • Start date Start date
  • Tags Tags
    Limit Sequence
Click For Summary
The discussion revolves around finding a sequence \( A_n \) such that \( \lim_{n \to \infty} \frac{A_n}{A_{n+1}} = \infty \) while ensuring \( \lim_{n \to \infty} A_n = \infty \). Participants suggest that setting \( \frac{A_{n+1}}{A_n} = n \) leads to the recursive relation \( A_{n+1} = n A_n \). It is concluded that sequences like \( A_n = C n^n \) or \( A_n = C n! \) can satisfy the condition, with the limit evaluated as \( e \cdot \infty = \infty \). The conversation emphasizes the need for a proper form of the sequence to achieve the desired limit behavior.
esuahcdss12
Messages
10
Reaction score
0
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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
24
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K