Two limits are the same

  • Thread starter Mr Davis 97
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In summary: Okay, that makes sense to me, and it helps me justify it intuitively in my head. I still have a concern though, specifically about the distinction between scenario 1) and scenario 2) that you described. Say in a proof that I wanted to justify my use of 1); I could just quickly note that ##a_{n+k}## is a subsequence of ##a_n## so that they have the same limit. But for 2), it doesn't seem I can quickly justify it, since ##a_{n-k}## is not a subsequence of ##a_n##. So how could I justify quickly that they have the same limit? Is there something distinct about showing 1) is true
  • #1
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Homework Statement


Let ##(a_n)## be a arbitrary real sequence. Given that the sequence ##\frac{a_{n+1}}{a_n}## is convergent, show that ##\lim \frac{a_{n+1}}{a_n} = \lim \frac{a_n}{a_{n-1}}##

Homework Equations


Take ##\mathbb{N} = \{1,2,3, \dots\}##

The Attempt at a Solution


In general, I know that if ##(b_n)## is any convergent sequence then the limit of any subsequence is the same as the limit ##(b_n)##. So, for example, ##\lim_{k \to \infty} b_{k+1} = \lim_{n \to \infty} b_n##. However, I am not sure how to apply this to my current problem. It doesn't seem like ##\frac{a_k}{a_{k-1}}## is a subsequence of ##\frac{a_{n+1}}{n}## since the selection function would have to be ##n_k = k-1##, and this is not a map from ##\mathbb{N}## to ##\mathbb{N}##, since ##n_1 = 0 \not \in \mathbb{N}##.
 
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  • #2
This is easy to prove by contradiction. Assume the limits of the two sequences are L1 and L2 and that the diff between them is h>0. Deduce a contradiction using epsilon-delta arguments. Then we must have h=0.
 
  • #3
Mr Davis 97 said:

Homework Statement


Let ##(a_n)## be a arbitrary real sequence. Given that the sequence ##\frac{a_{n+1}}{a_n}## is convergent, show that ##\lim \frac{a_{n+1}}{a_n} = \lim \frac{a_n}{a_{n-1}}##

Homework Equations


Take ##\mathbb{N} = \{1,2,3, \dots\}##

The Attempt at a Solution


In general, I know that if ##(b_n)## is any convergent sequence then the limit of any subsequence is the same as the limit ##(b_n)##. So, for example, ##\lim_{k \to \infty} b_{k+1} = \lim_{n \to \infty} b_n##. However, I am not sure how to apply this to my current problem. It doesn't seem like ##\frac{a_k}{a_{k-1}}## is a subsequence of ##\frac{a_{n+1}}{n}## since the selection function would have to be ##n_k = k-1##, and this is not a map from ##\mathbb{N}## to ##\mathbb{N}##, since ##n_1 = 0 \not \in \mathbb{N}##.
This seems to be quite obvious, since ##\lim_{k \to \infty} b_{k+1} = \lim_{k \to \infty} b_k## is all you need, just define ##b_k :=\dfrac{a_k}{a_{k-1}}##. You can use the definition of the convergence of one and take ##N+1## as the boundary for the ##\varepsilon## of the other.

More interesting would be, if ##\lim_{n \to \infty}\dfrac{a_n}{a_{n+1}}=\lim_{n \to \infty}\dfrac{a_{n+1}}{a_n}## or not.
 
  • #4
fresh_42 said:
This seems to be quite obvious, since ##\lim_{k \to \infty} b_{k+1} = \lim_{k \to \infty} b_k## is all you need, just define ##b_k :=\dfrac{a_k}{a_{k-1}}##. You can use the definition of the convergence of one and take ##N+1## as the boundary for the ##\varepsilon## of the other.

More interesting would be, if ##\lim_{n \to \infty}\dfrac{a_n}{a_{n+1}}=\lim_{n \to \infty}\dfrac{a_{n+1}}{a_n}## or not.
It's clear that ##\lim b_{k+1} = \lim b_k## because ##b_{k+1}## is a subsequence of ##b_k##, right? If that's so then how could I just as easily show that ##\lim b_k = \lim b_{k-1}##, since ##b_{k-1}## is not a subsequence of ##b_{k}##? How could I do this without going through a whole epsilon-delta proof?
 
  • #5
Mr Davis 97 said:
It's clear that ##\lim b_{k+1} = \lim b_k## because ##b_{k+1}## is a subsequence of ##b_k##, right? If that's so then how could I just as easily show that ##\lim b_k = \lim b_{k-1}##, since ##b_{k-1}## is not a subsequence of ##b_{k}##? How could I do this without going through a whole epsilon-delta proof?
If ##a_n## is a convergent sequence, then so is a sequence which starts with ##a_{n+k}## (1)
If ##a_n## is a convergent sequence, then so is ##b_1,\ldots ,b_k,a_1,a_2,\ldots## (2)

You can always either skip a finite number of sequence members or add them. This is even true, if it is not at the beginning of the sequence. As it is a finite number of additional or missing sequence elements, we can cut all out and start behind them with the new sequence.

##\forall \,\varepsilon>0 \,\exists \,N(\varepsilon)\in \mathbb{N} \,\forall \,n>N(\varepsilon)\, : \,|a_n-L|<\varepsilon##

So whether we cut finitely many sequence elements or add finitely many, say among the first ##M## elements, we simply choose ##N(\varepsilon)':=N(\varepsilon)+M## as our new boundary, and we will be on the safe side. That's the crucial point: We do not need to choose ##N(\varepsilon)## as low as possible, it doesn't matter, and if in doubt: double it, add a million or whatever.
 
  • #6
fresh_42 said:
If ##a_n## is a convergent sequence, then so is a sequence which starts with ##a_{n+k}## (1)
If ##a_n## is a convergent sequence, then so is ##b_1,\ldots ,b_k,a_1,a_2,\ldots## (2)

You can always either skip a finite number of sequence members or add them. This is even true, if it is not at the beginning of the sequence. As it is a finite number of additional or missing sequence elements, we can cut all out and start behind them with the new sequence.

##\forall \,\varepsilon>0 \,\exists \,N(\varepsilon)\in \mathbb{N} \,\forall \,n>N(\varepsilon)\, : \,|a_n-L|<\varepsilon##

So whether we cut finitely many sequence elements or add finitely many, say among the first ##M## elements, we simply choose ##N(\varepsilon)':=N(\varepsilon)+M## as our new boundary, and we will be on the safe side. That's the crucial point: We do not need to choose ##N(\varepsilon)## as low as possible, it doesn't matter, and if in doubt: double it, add a million or whatever.
Okay, that makes sense to me, and it helps me justify it intuitively in my head. I still have a concern though, specifically about the distinction between scenario 1) and scenario 2) that you described. Say in a proof that I wanted to justify my use of 1); I could just quickly note that ##a_{n+k}## is a subsequence of ##a_n## so that they have the same limit. But for 2), it doesn't seem I can quickly justify it, since ##a_{n-k}## is not a subsequence of ##a_n##. So how could I justify quickly that they have the same limit? Is there something distinct about showing 1) is true versus showing that 2) is true?
 
  • #7
Mr Davis 97 said:
Okay, that makes sense to me, and it helps me justify it intuitively in my head. I still have a concern though, specifically about the distinction between scenario 1) and scenario 2) that you described. Say in a proof that I wanted to justify my use of 1); I could just quickly note that ##a_{n+k}## is a subsequence of ##a_n## so that they have the same limit. But for 2), it doesn't seem I can quickly justify it, since ##a_{n-k}## is not a subsequence of ##a_n##. So how could I justify quickly that they have the same limit? Is there something distinct about showing 1) is true versus showing that 2) is true?
Simply cut off the first ##k## elements, they do not affect the convergence. You don't need the subsequence argument. Skip whatever is different and start at the point at which they are identical:

##(1)\, : \,1,2^{-1},3^{-1},4^{-1},5^{-1},6^{-1},\ldots##
##(2)\, : \,100, 200, 300,400,2^{-1}, 3^{-1},\ldots##
##(3)\, : \,6^{-1},7^{-1},8^{-1},\ldots##

For each sequence ##(i)## we have: ##\forall \,\varepsilon>0 \,\exists \,N_i(\varepsilon)\in \mathbb{N} \,\forall \,n>N_i(\varepsilon)\, : \,|a_n-L|<\varepsilon##.
So which sequence you choose, I define ##N(\varepsilon) := \max\{\,N_1(\varepsilon),N_2(\varepsilon),N_3(\varepsilon)\,\} + 9## and stop worrying about subsequences, which example you have chosen, or whether there are smaller values of ##N(\varepsilon)## possible. I don't care. And if you mess around with the first million sequence elements, I will define ##N(\varepsilon):=\max\{\,N_i\,\}+ 1,000,000##.

Now assume we know the convergence of sequence ##(i)## and want to prove the convergence for sequence ##(j)##. They are all equal from ##a_n=6^{-1}## on. This corresponds to ##n=6## for sequence ##(1)##, ##n=9## for sequence ##(2)## and ##n=1## for sequence ##(3)##. So the value ##N_j(\varepsilon)## can be chosen as ##N_j(\varepsilon)=N_i(\varepsilon)+9## since we know for sure that for all ##n>N_j(\varepsilon)## we have ##|a_n-L|<\varepsilon## from the convergence of sequence ##(i)##.
 
  • #8
Mr Davis 97 said:
It doesn't seem like ##\frac{a_k}{a_{k-1}}## is a subsequence of ##\frac{a_{n+1}}{n}##

Did you mean to write "##\frac{a_{n+1}}{a_n}##"?

If we are talking about the sequences denoted by ##\frac{a_n}{a_{n-1}}## and ##\frac{a_{n+1}}{a_n}## then what exactly does this notation mean? Using the convention that the first index of a sequence is 1, what are the first terms of each of those sequences? Are the two sequences the same sequence?
 

What does it mean when two limits are the same?

When two limits are the same, it means that the values of the two functions being evaluated as the limit approach the same value. In other words, the behavior of the two functions is identical as they approach a certain point.

How do you prove that two limits are the same?

To prove that two limits are the same, you must show that the values of the two functions being evaluated as the limit approach the same value. This can be done by using algebraic manipulation or by using the definition of a limit.

Can two limits be equal but the functions be different?

Yes, it is possible for two limits to be equal but for the functions to be different. This can happen when the behavior of the two functions is identical only at a certain point, but differ in other areas.

What is the significance of two limits being the same?

When two limits are the same, it means that the behavior of the two functions being evaluated is identical at that point. This can help in simplifying mathematical expressions and making predictions about the behavior of the functions.

Can two limits that are the same have different rates of change?

Yes, two limits that are the same can have different rates of change. The equality of limits only indicates that the behavior of the two functions is identical at a certain point, but does not determine the rate of change at that point.

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