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First prove $s_n$ converges iff $t_n$ converges

$t_n=s_{n+k}.$ prove $s_n \rightarrow s \iff t_n \rightarrow s$

This is obvious but I am not so sure how to write the proof.

Have:

$\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$

and

$N_2 = N_1+k$

$\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |t_n-t|<\varepsilon$

Not sure exactly how to turn this into a proof.

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    I answered my question below with a proof. Is it correct?2012-10-29

4 Answers 4

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You have to prove two implications: (1) if $s_n\to s$, then $t_n\to s$, and (2) if $t_n\to s$, then $s_n\to s$. I’ll take you through one of them carefully.

Start with what you’re assuming:

Assume that $s_n\to s$.

Use any relevant definitions to translate that into more fundamental terms:

Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|s_n-s|<\epsilon$ whenever $n\ge m_\epsilon$.

Now ask yourself just what it is you want to prove: you want to show that $t_n\to s$. Here again you should use any relevant definitions to reduce this to more basic terms: you want to show that for any $\epsilon>0$ there is a $k_\epsilon\in\Bbb N$ such that $|t_n-s|<\epsilon$ whenever $n\ge k_\epsilon$. To do this, you’ll start with any old $\epsilon>0$ and show how to find such a $k_\epsilon$, so the next step of the proof must be:

Let $\epsilon>0$; we must show that there is some $k_\epsilon\in\Bbb N$ such that $|t_n-s|<\epsilon$ whenever $n\ge k_\epsilon$.

(You don’t actually need the part after the semicolon, but it doesn’t hurt, especially when you’re just beginning.)

We want to get $|t_n-s|$ small, and we know how to make $|s_n-s|$ small, so clearly we should see exactly how $t_n$ is related to $s_n$. That’s easy: $t_n=s_{n+k}$.

Suppose that $n\ge m_\epsilon$; then $n+k\ge m_\epsilon$, so $|t_n-s|=|s_{n+k}-s|<\epsilon$. Thus, if we let $k_\epsilon=m_\epsilon$, we have $|t_n-s|<\epsilon$ whenever $n\ge k_\epsilon$, and it follows that $t_n\to s$.

Putting everything together in one place:

Assume that $s_n\to s$. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|s_n-s|<\epsilon$ whenever $n\ge m_\epsilon$. Let $\epsilon>0$; we must show that there is some $k_\epsilon\in\Bbb N$ such that $|t_n-s|<\epsilon$ whenever $n\ge k_\epsilon$. Suppose that $n\ge m_\epsilon$; then $n+k\ge m_\epsilon$, so $|t_n-s|=|s_{n+k}-s|<\epsilon$. Thus, if we let $k_\epsilon=m_\epsilon$, we have $|t_n-s|<\epsilon$ whenever $n\ge k_\epsilon$, and it follows that $t_n\to s$.

You can use the same approach to prove (2). You’ll have to work a little harder, but only a little; feel free to ask questions if you get completely stuck.

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    I attempted t$h$e proof below, can you see if the part 2 is correct? I had to avoid n<1 and I am not sure I did it correctly.2012-10-29
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The forward direction follows from the fact that every subsequence of a convergent sequence converges to the same limit(s).

For the backward direction, suppose that $t_n\to s$. Take $\epsilon>0$ arbitrary, so by assumption, there is some $N$ such that $|t_n-s|<\epsilon$ for $n\geq N$. Putting $N'=\max\{1,N-k\}$, we have for $n\geq N'$ that $|s_n-s|=|t_{n+k}-s|<\epsilon$, and so $s_n\to s$ by definition.

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Attempt at proof:

For all $ \varepsilon >0$, there exists $ N\in \mathbb N$ such that for all $n\in \mathbb N,\,n>N,\, |s_n-s|<\varepsilon$

For $n + k> n >N$, $|t_n - s| = |s_{n+k} - s |<\varepsilon$.

So $s_n \rightarrow s$ implies $t_n \rightarrow s$.

and then the other way:

For all $\varepsilon >0$, there exists $M\in \mathbb N$ such that for all $n\in \mathbb N,\,n>M,\, |t_n-t|<\varepsilon$

Let $N = \max\{M-k,1\}$. For $n>N$, $|s_{n}-t|=|t_{n-k }- t| < \varepsilon$.

So So $t_n \rightarrow t$ implies $s_n \rightarrow t$. Which completes the proof.

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    Edited, how is it now?2012-10-29
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One direction is immediate.

Suppose we assume that $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$. So, choose $\epsilon>0$ and let $N_1$ be the number given by our assumption. Then if $n> N_1$, obviously you have $n+k > N_1$, and hence $|s_{n+k}-s|<\varepsilon$ or in other words, $|t_n-s|<\varepsilon$. Hence, under the given assumption we have $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |t_n-s|<\varepsilon$.

(Note that the particular label for $N_2$ doesn't matter, we could equally well have written $\forall \varepsilon >0,\,\exists \aleph \in \mathbb N$ such that $\forall n\in \mathbb N,\,n>\aleph,\, |t_n-s|<\varepsilon$, or any other unambiguous symbol we want. The same applies to all symbols used, modulo readability.)

Now suppose $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |t_n-s|<\varepsilon$. This can be rewritten as $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |s_{n+k}-s|<\varepsilon$. As usual, let $\epsilon >0$. Now choose $N_1 = N_2+k$, and note that if $m>N_1$, we have $m-k >N_2$ and hence $|s_m -s| < \varepsilon$. Hence we have $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,m>N_1,\, |s_m-s|<\varepsilon$, or in the symbols you used initially, $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$.