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To prove that a sequence converges, find one or more convergent subsequences

Quick description

To prove that a sequence converges, it is sometimes easier to start by finding a subsequence that converges (or proving that such a subsequence exists). Then one can hope to deduce that the sequence itself converges. Another way of using subsequences is to exploit the following result: if every subsequence has a further subsequence that converges to a limit L, then the whole sequence converges to L.

Prerequisites

Basic concepts of real analysis.

Note iconIncomplete This article is incomplete. Some general discussion would be nice, as would an example of the second method of proving convergence.

Example 1

To prove that a Cauchy sequence (x_n) of real numbers converges, assuming one of the other forms of the completeness axiom, one can argue as follows. First, one proves the Bolzano-Weierstrass theorem. Then one observes that a Cauchy sequence is bounded, and so by the Bolzano-Weierstrass theorem has a convergent subsequence. And finally, one proves that if a Cauchy sequence has a subsequence that tends to a limit L, then the sequence itself tends to L.

Example 2

Here is a second way of proving that Cauchy sequences converge. Let (x_n) be a Cauchy sequence. By the Cauchy condition we can choose a subsequence (x_{n_k}) such that |x_{n_k}-x_m|\leq 2^{-k} whenever m\geq n_k. It follows that the closed intervals [x_{n_k}-2\cdot 2^{-k},x_{n_k}+2\cdot 2^{-k}] are nested, since if k<l then x_{n_l}\in[x_{n_k}-2^{-k},x_{n_k}+2^{-k}], and 2\cdot 2^{-l}<2^{-k}. Therefore, their intersection is non-empty. Since their lengths tend to zero, the intersection is a singleton \{x\}, and it is easy to check that the subsequence x_{n_k} converges to x. Therefore, since (x_n) is Cauchy, so does the whole sequence.

General discussion

The trick in the second approach was to replace a Cauchy sequence by a subsequence that was rapidly Cauchy. That gave us the inequality 2\cdot 2^{-l}<2^{-k}, which was crucial to the success of the proof.