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No explanation required for the question I guess.

Picasso
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    The statement doesn't really make any sense; unless, of course, you're using a different definition. Convergence is defined as $\forall; \varepsilon > 0, \exists N$ such that $\forall; n>N, |a_{n}-L|<\varepsilon$. Divergence is the opposite. I guess by an empty argument, the sequence would converge to its last element? – Christopher K Dec 10 '13 at 02:29
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    @ChrisK can't you say that 1,1,1,1,1,.... (n terms) converges to 1 ? – Bach Dec 10 '13 at 02:34
  • @Bach, perhaps. I've never seen convergence used in this context; it seems like a very vacuous argument to me though. – Christopher K Dec 10 '13 at 02:37
  • @StefanSmith : I was just thinking about it myself. And no I do not mean an infinite sequence with a finite number of distinct elements. – Picasso Dec 11 '13 at 14:32
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    Generally the word "sequence" means infinite sequence. – Jair Taylor Dec 22 '22 at 05:05

1 Answers1

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Yes. A finite sequence is convergent.

Call your sequence $\{a_k\}$. It is finite, so it has a last term, say $a_m=M$.

An sequence converges to a limit $L$ if for any $\epsilon>0$, there exists some integer $N$ such that if $k\ge N$, $|a_k-L|<\epsilon$. However, since your sequence is finite, for any $\epsilon>0$ we just take $N=m$, and it is clearly true that if $k\ge m$, $|a_k-L|=|a_m-L|=0<\epsilon$, since the only possible value for $k$ is $M$ itself.

I hope that was not too confusing.

ant11
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    I don't think that this application of the definition is correct since when you write $|a_k-L|$ for all $k\ge m$, you are implicitly assuming that $a_k$ for all $k\ge m$ does make sense, which is not true. – Atom Apr 14 '21 at 17:11