2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '16 at 22:53
  • $\begingroup$ 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/… $\endgroup$ – John Math Aug 9 '16 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. $\endgroup$ – Jacob Wakem Aug 10 '16 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. $\endgroup$ – Jacob Wakem Aug 11 '16 at 0:50
  • 1
    $\begingroup$ Your statement only holds when a_0 exists! $\endgroup$ – Jacob Wakem Oct 28 '16 at 0:02
2
$\begingroup$

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 $(a_{n-1}) \to L$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Benedict Voltaire Aug 9 '16 at 23:00
  • 1
    $\begingroup$ 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 '16 at 23:03
0
$\begingroup$

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 $|a_n - L|<\epsilon$.

$\endgroup$
0
$\begingroup$

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 $L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.