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How do you prove that given a sequence that is finite, the sequence always converges? Or, must the sequence be cauchy for this finite sequence to converge? How can a cauchy sequence be finite? If a cauchy sequence is finite, won't all the elements in the sequence be arbitrarily close?

darkgbm
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  • It does not make sense to say that a finite sequence converges. And if you take the definition in Wikipedia, all finite sequences are Cauchy. http://en.wikipedia.org/wiki/Cauchy_sequence –  Oct 11 '14 at 15:45

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Let $\{ a_k \}_{k=1}^{n}$ be a finite sequence. This sequence converges to $a_n$, because for all $k \geq n \in \mathbb{N}$, $|a_k - a_n| = |a_n - a_n| = 0 < \epsilon$ for any $\epsilon > 0$.

It's not a very interesting case, but a good problem for understanding the definition of convergence.

Bruce Zheng
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  • I like your reasoning. There's only one thing that troubles me: The formal definition of a sequence $(a_k)$ being convergent to $a$ says that $$ (\forall \epsilon > 0)(\exists N \in \mathbb{N}) k > N \rightarrow |a - a_k| < \epsilon$$ This means that the implication also has to hold for k > n. But $a_{n+1}$ is undefined.. – Max Herrmann Jun 07 '15 at 18:56
  • Probably too old but for future people... any truly 'finite sequence' can't be cauchy or convergent, since supposing k>N satisfied for N being the amount of terms in the finite sequence, will mean that the consequent will be false, as the left side of the inequality to epsilon won't be defined, making the inequality false. Instead, we have to consider an infinite sequence, with the 'finite' meaning the RANGE of the sequence is bounded, that is, a difference from {1} and {1,1,1...}. The latter is convergent to 1, the former isn't, despite having the same range. – VgAcid Aug 17 '18 at 05:15
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Since the given sequence is finite, there is a term say $b$ which repeats infintely many times i.e there exists an $n_0$ such that for all $n \ge n_0$, we have $a_n=b$. And the consequences follow

tattwamasi amrutam
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