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Is there a quick and easy way of proving that if $\|x_{n+1} - x_n\|\to 0$ and $\|x_n\|< B$ then $x_n$ converges?

If $\|x_n\|< B$, then there is $x_{n_k}\to x^*$ (by Bolzano Weierstrass). Then I thought maybe I could write $$ \|x_n - x^*\| < \|x_n - x_{n-1}\| + \cdots + \|x_{n_k + 1} - x_{n_k}\| + \|x_{n_k} - x^*\| $$ where $n_k$ is the largest element of the subsequence that is less than $n$. The problem is that when I try to bound this, the $\epsilon$ is kind of "dependent" on $n$, which isn't the statement we're trying to prove.

So, I'm a little stuck. I haven't tried assuming $x_n$ doesn't converge for a contradiction, so I'll try that while this question is up. Is there a nice quick way to prove what I'm looking for?

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The given statement is false.

Consider $\{x_n\} = \{0,1,1/2,0,1/4,1/2,3/4,1,7/8,3/4,5/8,\ldots\}$.

This sequence is bounded, and clearly not convergent (there are obvious subsequences that are constant with value 0 and 1), and by construction $\|x_n - x_{n+1}\| \to 0$ as $n \to \infty$.

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  • $\begingroup$ Damn, I had a suspicion. Thanks for coming up with a counterexample! $\endgroup$ Feb 12 '18 at 3:23

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