# 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
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?