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I haven't dealt with convergence of finite sequences yet and my textbook doesn't say much about this. Using the definition of convergence, I was able to show that the limit of a finite sequence is its last term. I don't know if this is correct. Can you please let me know?

Let $\{a_n \}_{n \in N}$ be a finite sequence where $N \subset \mathbb{N}$. Denote the number of elements in $N$ by $\bar{N}$. Now we know that for all $\epsilon > 0, \ \exists \ \bar{N} \in \mathbb{N}$ such that if $ n \geq \bar{N}, \ \mid a_\bar{N}-a_n \mid<\epsilon$. But this is the definition of a sequence that converges to $a_\bar{N}$. So does this mean that the limit of a finite sequence is simply the last element of the sequence?

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  • $\begingroup$ And by this logic, is the limit of a sequence with one element $\{a_1\}$ just $a_1$? $\endgroup$
    – lasoon
    Apr 7 '17 at 23:21
  • $\begingroup$ Seems quite unusual to take the limit of a finite sequence. But the definition you mentioned seems to be conform to the definition of a limit $\endgroup$
    – Peter
    Apr 7 '17 at 23:23
  • $\begingroup$ I'm not voting to close, but note this is a duplicate of math.stackexchange.com/questions/600795/… $\endgroup$ May 27 '17 at 4:21
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Technically yes, I think, but it is not too interesting to consider convergence of a sequence that terminates after finitely many terms.

If $a_1,...,a_{n_0}$ is a finite sequence indexed by $I = \{1,...,n_0\}$ and $x = a_{n_0}$, then the criterion for convergence $a_n \to x$ is automatically satisfied: $$(\forall \varepsilon > 0)(\exists N_0 \in I)(n \geqslant N_0, \, n \in I \,\, \Longrightarrow \,\,|a_n-x|<\varepsilon).$$ No matter what $\varepsilon$ you pick setting $N_0 = n_0$ will work.

Typically the index set is $I = \mathbb{N}$.

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  • $\begingroup$ Yes, I agree. I also observed that setting $N_0=n_0$ does the trick for all $\epsilon>0$, but I wasn't sure if I was misunderstanding something. Thanks for confirming. $\endgroup$
    – lasoon
    Apr 7 '17 at 23:46
  • $\begingroup$ Is there anything wrong with using this result to prove that any bounded sequence has a convergent subsequence? One could simply define a subsequence of the convergent sequence to contain one element. Then, by the same argument we have been making, this subsequence with one element converges to the one element, and thus, i have shown that a bounded sequence has a convergent subsequence (although this could also work for an unbounded sequence). Is there a flaw in my logic? I think there is. I just don't know what it is. Or is the question poorly specified, thereby, allowing this to be a proof. $\endgroup$
    – lasoon
    Apr 8 '17 at 0:02
  • $\begingroup$ If you are referring to the Bolzano-Weierstrass theorem, then I don't suggest you use finite sequences. That isn't really what the theorem is asserting anyway: when the theorem says subsequence it is understood to mean infinite subsequence. Knowing that there are finite subsequences that converge tells you pretty much nothing about the space you are studying. Knowing the existence of infinite subsequences tells you much more $\endgroup$
    – joeb
    Apr 8 '17 at 0:20
  • $\begingroup$ Ok. Thank you. I will work out a proof for infinite subsequences. $\endgroup$
    – lasoon
    Apr 8 '17 at 0:26
  • $\begingroup$ No. If the sequence is $a = (1,2,3,4)$ and $N_0 = 1$, then $10 > N_0$ but $a_{10}$ has no meaning. $\endgroup$ Jan 31 '18 at 9:49
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The limit is usually only defined for an infinite sequence, not a finite sequence.

The usual definition:

For all $\epsilon > 0$ there is $N$ such that for every $n > N$, $|a_n - L| < \epsilon$

assumes $a_n$ to be defined for every $n > N$, which is not the case for a finite sequence.

If you modify it to say "... for every $n > N$ for which $a_n$ is defined, ...", then the limit of a finite sequence could be anything at all. There's no particular reason for it to be the last member of the sequence: $L = $"a unicorn" works just as well if $N$ is greater than the index of the last member of the sequence.

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  • $\begingroup$ I don't see why your argument in the last paragraph holds. Also, the definitions I have come across are different than yours. They usually read : For all $\epsilon>0$ there is $N$ such that if $n \geq N$, $\mid a_n-L \mid < \epsilon$. $\endgroup$
    – lasoon
    Apr 7 '17 at 23:41
  • $\begingroup$ @lasoon are you familiar with vacuuously true statemenst. Le $k =$ the number of terms in the sequence. Then let $N > k$. Then for all $n > N$ it is true that $|a_n - 57.39| < \epsilon$ because there is no $a_n$. ... $\endgroup$
    – fleablood
    Apr 7 '17 at 23:47
  • $\begingroup$ @fleablood Yes, but it doesn't necessarily hold if you set $N=k$ and take $\mid a_n - 57.39 \mid$ for $n \geq N$. The definition in my post is analogous to letting $N=k$ and taking the difference for $n \geq N$. Is my definition wrong? Should $n \geq N$ be switched with $n>N$ and $N = k$ be switched with $N > k$ ? $\endgroup$
    – lasoon
    Apr 7 '17 at 23:54
  • $\begingroup$ It doesn't matter whether you use $\ge$ or $>$. $n > N$ is the same as $n \ge N+1$ (for integers). $\endgroup$ Apr 8 '17 at 0:08
  • $\begingroup$ On the other hand. The "for any n" refers to the n, not the a_n. so "there is no a_n" could go either way. Frankly I don't think it mayters. $\endgroup$
    – fleablood
    Apr 8 '17 at 3:23

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