If the sequence $(a_n)_{n \in \mathbb {N}} = (a_n)$ converges to the real number $L $, then the sequence $(a_{n-1})_{n \in \mathbb {N}} = (a_{n-1}) $ also converges to $L $.

This appears true intuitively, but I'm trying to write a proof.

Suppose $(a_n) $ converges to $L $. Then for every $\epsilon > 0$, there is an $N \leq n $ such that $a_n \rightarrow L $. Choose $N=n-2$. Then $N < n-1$, and since the chosen $\epsilon $ corresponds with the selected $N $ (in reference to the order of the quantifiers in the definition), it follows that $a_{n-1} \rightarrow L $.

Is this sort of correct? Is there a clearer way using the definition of the limit of a sequence?

  • $\begingroup$ I think the idea you're getting at is: if you're within $\epsilon$ distance of $L$ for all $n \geq N$ with $(a_n)$, then the same is true for $(a_{n-1})$ but with $N+1$. More generally, any subsequence of a convergent sequence converges to the same limit. $\endgroup$
    – Kaj Hansen
    Aug 9, 2016 at 22:53
  • 1
    $\begingroup$ Have a look to the demonstration of "If a sequence converges, then every subsequence converges to the same limit.". It show you the way math.stackexchange.com/questions/213285/… $\endgroup$
    – Jean G
    Aug 9, 2016 at 22:58
  • $\begingroup$ In your definition you can replace (all large n) with (all but finitely many n). This shift should make the problem trivial. $\endgroup$ Aug 10, 2016 at 0:35
  • $\begingroup$ Please read my answer. It is the most obvious and straightforward attack. It shows the demonstrative power of the image of the sequence. $\endgroup$ Aug 11, 2016 at 0:50
  • 1
    $\begingroup$ Your statement only holds when a_0 exists! $\endgroup$ Oct 28, 2016 at 0:02

3 Answers 3


First of all your definition of convergence is kind of off:

$(a_n)$ converges to $L$ if and only if for all $\epsilon>0$ there is an $N \in \mathbb{N}$ such that $n>N$ implies $|a_n-L|<\epsilon$.

Now suppose $(a_n)$ converges to $L$. Fix $\epsilon>0$. Then there is an $N \in \mathbb{N}$ such that $n>N$ implies $|a_n-L|< \epsilon$. Then for this same $N$, $n-1>N$ implies $|a_{n-1}-L|<\epsilon$, hence $(a_{n-1}) \to L$.

  • $\begingroup$ That's the definition I prefer to use, but in the book I'm using (as well as Bloch' s Proofs and Fundamentals), they use the nonstrict inequality $N\leq n $, which is what's giving me issues. $\endgroup$
    – J. Dunivin
    Aug 9, 2016 at 23:00
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    $\begingroup$ You can use the nonstrict inequality if you want, it doesn't really make a difference. I think in the definition you give in the question you can't just say "there is an $N \leq n$" you have to say: "there is an N such that $n \geq N$ implies ..." $\endgroup$
    – M10687
    Aug 9, 2016 at 23:03

I don't think you've used the definition of the convergence of a sequence correctly, and in fact I'm not sure I understand what you mean by this:

"Then for every $\epsilon > 0$ there is an $N\leq n$ such that $a_n\rightarrow L$"

I think what you mean to say is the for any $\epsilon >0$, there is an $N$ such that if $n \geq N$, then $|a_n - L|<\epsilon$.


Definition: the $N$-th tail of a sequence is obtained by dropping the first $(N-1)$ terms of the sequence.

You should prove the following statement, which is more general, but at the same time much simpler, than your problem:

A sequence converges to $L$ if and only if its every tail converges to $L$.


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