Is every sequence that converges to 0 also has a convergent subsequence?

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The discussion centers on the convergence of sequences, specifically whether a sequence \( a_n \) converging to 0 implies that every subsequence also converges to 0. The participants confirm that if every subsequence converges to 0, then the sequence \( a_n \) itself must converge to 0. They reference the definition of convergence, using the triangle inequality to establish bounds on the sequence's behavior. The conversation also touches on the converse of the statement, prompting further exploration of counterexamples.

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Homework Statement


If every subsequence of the sequence a_n converges to 0, prove that a_n itself converges to 0. Is the converse true? Prove it or give a counter example.


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The Attempt at a Solution


I can prove a sequence is convergent if it has a convergent subsequence. What about every sequence is convergent?
 
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This is a similar reasoning to the proof that if xn converges to x, then all subsequences that converge do so to x. You can do this by considering subsequences xj and xk where j is odd and k is even.
 
Look at the definition of convergence. The one I learned was that a_n converges to a number L if and only if given e > 0 there exists an N such that |a_n - L| < e for all n>N. Hint: use the triangle inequality:

[tex]a_n - L = a_n + \underbrace{(- b_j + b_j - b_k + b_k)}_{=0} - L<br /> = (a_n - b_j) + (b_j - b_k) + (b_k - L)[/tex]

which implies

[tex]|a_n-L| \le |a_n - b_j| + |b_j - b_k| + |b_k - L|[/tex]

If you consider the b_j and b_k as subsequences, you should be able to bound the right hand side using the definition of convergence.
 

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