Question

Let \(X_n\) be a Cauchy sequence. Suppose that for every \(\varepsilon >0\), there is some \(n>1/\varepsilon\) such that \(|X_n|<\varepsilon\). Prove that \(X_n\rightarrow 0\)
Please, help.

Answer 1

My attempt:
Since \(X_n\) is a Cauchy sequence, Let \(\varepsilon >0\), then for m >n, there exists \(N_1 >0\) such that \(\forall~~m, n>N_1\) \(|X_n-X_m|< \varepsilon/2\)
And we have \(|X_n|= |X_n -X_k+X_k|\leq |X_n-X_k|+|X_k|\)
By above, if k > n, then the first term from the right hand side \(<\varepsilon/2\)
All we need is manipulate \(|X_k| <\varepsilon /2\). I stuck here. I don't know how to argue.

Answer 2

I know I need play around with the condition \(n > 1/\varepsilon\) such that \(|X_n|<\varepsilon\) to get what I want, but I don't know how.
Someone helps me, please

Answer 3

Why not just choose \(\varepsilon\) small enough so that \(\frac{1}{\varepsilon} > N_1\)?

Answer 4

Suppose you chose \(\varepsilon\), then using the condition choose:
\(\displaystyle\varepsilon'= \min\ \left\{\frac{1}{N_1}, \frac{\varepsilon}{2}\right\}\)

Then you have \(\displaystyle|X_n - X_k| + |X_k| < \frac{\varepsilon}{2} + \varepsilon' < \varepsilon\)

Answer 5

Got it, thanks a lot. :)