# Two limits are the same

## 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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andrewkirk
Homework Helper
Gold Member
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.

fresh_42
Mentor

## 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.

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?

fresh_42
Mentor
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.

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?

fresh_42
Mentor
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)##.

Stephen Tashi