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Suppose that the sequence $\{a_n\}$ converges to $l$ and that sequence $\{b_n\}$ has the property that there is an index $N$ such that $\{a_n\}=\{b_n\}$ for all $n\geq N$. Show that $\{b_n\}$ converges to $l$.

So far I have used the definition of convergence on $\{a_n\}$. So $|a_n-l|<\epsilon$. Then I thought to use the Comparison Lemma where $C=1$, so we get $|b_n - l| \leq |a_n-l|<\epsilon$, which is true since $\{a_n\}=\{b_n\}$ for all $n\geq N$. That would clearly show what I needed to prove, but the problem is that I feel the last step is too much of a "leap of faith."

Any help is grateful, and thank you in advance.

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If $n \ge N$, then $a_n-l=b_n-l$. Pick any $\epsilon>0$. Let $M$ be such that $|a_n-l|<\epsilon$ if $n >M$. (There is such an $M=M_{\epsilon}$ by the definition of convergence.) If $P=\max(N,M)$ then if $n>P$, we have $|b_n-l|=|a_n-l|<\epsilon$. –  André Nicolas Sep 27 '11 at 21:58
    
What is the comparison lemma? (I haven't heard of this used in the context of taking limits.) –  Srivatsan Sep 27 '11 at 21:59
    
You are right in your proof. Just the inequality should be forall $n \geq N$. It seems trivial because the property is obvious. –  Beni Bogosel Sep 27 '11 at 22:03
    
@SrivatsanNarayanan: I don't think that in English this is stated as comparison lemma. In Romania, that's the name. The theorem says that if $a_n$ has the property that there exists $\ell$ and $b_n \geq 0$ with $b_n \to 0$ and $|a_n-\ell| \leq b_n$ then $a_n \to \ell$. Maybe it doesn't have a name in English literature –  Beni Bogosel Sep 27 '11 at 22:07
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2 Answers

up vote 3 down vote accepted

It is straight from the definition.

The convergence of a sequence does not depend on its first $k$ terms, where $k$ is any natural number. It's "the tails" that matter.

If $a_n$ and $b_n$ are equal from the $N$-th term on, you can define a sequence $c_1=a_{N}$, $c_2=a_{N+1}$ and so on, and this sequence has the same limit as $a_n$ and $b_n$. Because, given $\varepsilon > 0$, you know that there exists $M$ such that for $n\geq M$, $|l-a_n|<\varepsilon$. But then, all the more so, for $n\geq M$, $|l-c_n|<\varepsilon$.

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The definition says that $a_n \to x$ if for every $\varepsilon > 0$ there exists $n_0 \in \Bbb{N}$ such that $|a_n-x| <\varepsilon \forall n \geq n_0$.

Now, if $b_n=a_n$ for every $n \geq N$, then taking $M=\max\{n_0,N\}$ you have that $|b_n-x| < \varepsilon, \ \forall n \geq M$, and by definition $b_n \to x$.

You can use comparison lemma like this.

$|b_n-x| \leq (=) |a_n-x|, \ \forall n\geq N$. Taking $n \to \infty$ we get that $|b_n-x| \to 0$ and $b_n \to x$.

Another heuristic argument would be that taking the limit means studying the sequence as $n$ is close to $\infty$. Therefore, if two sequences are equal from a point on, then their behavior when $n$ is big enough is the same. Maybe that's why any of the above proofs seems like a tautology or stating something obvious.

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