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The full question wouldn't fit in the title. Here it is:

Let ($a_n$)$_n$$_\in$$_\Bbb{N}$ and ($b_n$)$_n$$_\in$$_\Bbb{N}$ be two sequences and suppose that the set {n$\in$$\Bbb{N}$: $a_n\neq b_n$} is finite ($a_n$ and $b_n$ differ for finitely many values of n). Prove that either both sequences converge to the same limit or both diverge.

Our definition for convergence is: Given a real number $L$, we say that $(X_n)$ converges to L if for every $\epsilon$>0, there exists N∈$\Bbb{N}$ such that for all n∈N satisfying n>N, we have |$X_n-L$|<$\epsilon$. I'm just not sure if I'm supposed to apply that definition to this problem and if so, then how?

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    Hint: take $N = 1+\max\{n\in\Bbb{N}: a_n\neq b_n\}$ – 2017-02-20
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    There is such an N iff all but finitely many values are within epsilon of L iff all but finitely many values of a sequence that is identical except at finitely many values is within epsilon of L. – 2017-02-20
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    Their difference converges to 0. This is sufficient. – 2017-02-21

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If the set $\{n\in \mathbb N: a_n \neq b_n\}$ is finite, there is an $N_0$ such that $a_n=b_n \ \forall n>N_0$. Now, suppose that $\lim_{n} a_n=L$ exists and it's a real number. Then,

$$\forall \varepsilon>0 \quad \exists N\in \mathbb N: n>N \Rightarrow |a_n-L| <\varepsilon$$ and now we'd like to prove that $\lim_{n} b_n=L$, that is, $$\forall \varepsilon>0 \quad \exists M\in \mathbb N: n>M \Rightarrow |b_n-L| <\varepsilon$$ but if we choose $M=\max\{N, N_0\}$, then $$n>M \Rightarrow |b_n-L|=|a_n-L|<\varepsilon$$

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    How are you saying that if the set is finite, then there is an $N_0$ such that $a_n$=$b_n$? Also, what do you mean by 0$\exists$$M$? – 2017-02-20
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    It is a typo. It has to be $a_n \neq b_n$. – 2017-02-20
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    @johnie4usc reasoning by contradiction, if for every $n\in \mathbb N$, there is $k(n)>n$ such that $a_{k(n)}\neq b_{k(n)}$, we have that $k(1), k(2), k(3)...$ are indices for which $a_{k(n)}\neq b_{k(n}$, and there are as many as the natural numbers. That contradicts the finiteness of the set. – 2017-02-20
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    @NiklasHebestreit no, it's not. If $a_n\neq b_n$ for finite terms, that set of indices can't have indices as big as you want. – 2017-02-21
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    @flytothesurface Yeah, it is late ... Sorry for that. – 2017-02-21
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    @NiklasHebestreit no problem at all~ – 2017-02-21
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    @flytothesurface okay. You're basically saying that since $\Bbb{N}$ is infinite and the set {n$\in$$\Bbb{N}$ : $a_n$ $\neq$ $b_n$} is finite, that there must exist an $N_0$ such that $a_n$ = $b_n$? – 2017-02-21
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    @flytothesurface I'm also getting confused about your n<$N$. Why would that be true? The definition of convergence is that n>$N$? – 2017-02-21
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    Yes, for all $n>N_0$, $a_n=b_n$. The $nN$. Sorry about that. – 2017-02-21
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    Okay, so are you saying they converge to the same limit? – 2017-02-21
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    Yes, that's it, because the sequences are equal, except for the first $N_0$ terms. – 2017-02-21
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    Okay. Let me write this all out in english and see if I'm understanding. Let $a_n$ and $b_n$ be sequences and suppose that the set {n$\in$$\Bbb{N}$: $a_n$$\neq$$b_n$} is finite. Then, there exists an $N_0$ such that $a_n$=$b_n$. Suppose the limit of $a_n$ = L. Then for each $\epsilon$>0, there exists an N$\in$$\Bbb{N}$ such that if n>N then |$a_n$ - L| < $\epsilon$. To show that the limit of $b_n$ = L, we must show that for all $\epsilon$ > 0, there exists an M$\in$$\Bbb{N}$ such that if n>M then |$b_n$ - L| < $\epsilon$. Let M=max{N, $N_0$}, then n>M and |$b_n$ - L| = |$a_n$ - L| < $\epsilon$ – 2017-02-22
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    I'm confused about a couple parts. First, how do you know that for all n, n>$N_0$? And why are you choosing the max{N, $N_0$} for M? – 2017-02-22
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    Could we just get rid of the n>$N_0$ and then when choosing M, we break down into cases such that M=N or M=$N_0$? Either way, the limit is the same. – 2017-02-22
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There exists an N for {a_n} iff there exists an N larger than max{n:a_n is not b_n} for {a_n} iff there exists an N for {b_n}.