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

In summary: 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.
  • #1
esuahcdss12
10
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
  • #2
I'm assuming you mean to study $\frac{A_n}{A_n + 1}$. Note that $\lim_{n \to \infty}\frac{A_n}{A_n + 1} = \lim_{n \to \infty}\frac{1}{1 + \frac{1}{A_n}}$.
 
  • #3
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))
 
  • #4
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)
 
  • #5
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?
 
  • #6
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)
 
  • #7
Another example would be:

\(\displaystyle A_{n}=Cn!\)

So that:

\(\displaystyle \frac{A_{n+1}}{A_{n}}=\frac{C(n+1)!}{Cn!}=n+1\)
 
  • #8
MarkFL said:
Another example would be:

\(\displaystyle A_{n}=Cn!\)

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.
 

1. What does "lim(An/An+1)=∞" mean?

This notation represents the limit of a sequence, where the n-th term divided by the (n+1)-th term approaches infinity. In other words, as the sequence progresses, the terms become infinitely larger in relation to each other.

2. How do you find a sequence to make lim(An/An+1)=∞?

To find a sequence that satisfies this condition, you need to choose a starting value for An and then make each subsequent term larger than the previous one. For example, you can choose An = n, so that An+1 = n+1, An+2 = n+2, and so on. This will result in a sequence where the terms increase infinitely as n approaches infinity.

3. Is this sequence unique?

No, there can be multiple sequences that satisfy the condition of lim(An/An+1)=∞. For instance, instead of choosing An = n, you can also choose An = 2n, which will also result in a sequence where the terms increase infinitely.

4. Can this limit be negative?

No, since the denominator of the fraction is (An+1), the limit can only approach positive infinity. If the limit were to approach negative infinity, the denominator would need to be (An-1).

5. What are some real-life applications of this concept?

The concept of finding a sequence to make lim(An/An+1)=∞ is used in various fields of science and engineering, such as in the study of growth rates in biology, the analysis of infinite series in mathematics, and the design of algorithms in computer science. It is also used in economics to model situations where quantities increase without bound.

Similar threads

Replies
11
Views
1K
  • Calculus
Replies
7
Views
2K
Replies
11
Views
1K
Replies
3
Views
899
Replies
2
Views
828
  • Calculus
Replies
1
Views
1K
Replies
3
Views
1K
  • Calculus
Replies
11
Views
1K
Back
Top