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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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Forgot to add that this is homework –  fhyve Oct 29 '12 at 5:40
    
For a bi-conditional/if-and-only-if statement make sure that you prove both directions: $P \Rightarrow Q$ and $Q \Rightarrow P$. –  BobaFret Oct 29 '12 at 5:44
    
I answered my question below with a proof. Is it correct? –  fhyve Oct 29 '12 at 6:20

4 Answers 4

up vote 1 down vote accepted

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 the 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. –  fhyve Oct 29 '12 at 6:25

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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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 $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$.

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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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1  
A proof is a piece of prose, consisting of paragraphs of sentences. Look at the final version of my proof of one direction: yes, it has a lot of mathematical notation in it, but it can be read aloud much as one would read a story or a newspaper article aloud. What you’ve written here is a bunch of disconnected bits and pieces, lacking almost all of the verbal connective tissue that would give it structure and make it both readable and understandable. That’s the sort of writing that you should use as a model; @copper.hat’s answer is another. –  Brian M. Scott Oct 29 '12 at 6:32
    
Once it’s properly organized and expanded to something readable, your first argument is fine, but it’s unnecessary: that’s the direction that I proved for you. You’re also working harder than necessary: if $n>N$, then $n+k>N$, so $N$ works for $\langle t_n:n\in\Bbb N\rangle$ as well as for $\langle s_n:n\in\Bbb N\rangle$. For the other direction you do have to make an adjustment, but you’ve made it in the wrong direction: you have $|t_n-s|<\epsilon$ when $n>M$, so $|s_{n+k}-s|<\epsilon$ when $n>M$, and $|s_n-s|<\epsilon$ when $n>M+k$. –  Brian M. Scott Oct 29 '12 at 6:38
    
Edited, how is it now? –  fhyve Oct 29 '12 at 8:16

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