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That is, let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be sequences such that $b_n \overset{n \to \infty}{\to} 0$ and for all $k \in \mathbb{N}$ and $l \geq k$, $$|a_l - a_k| < b_k\text{.}$$ Show that $(a_n)_{n=1}^{\infty}$ is Cauchy.

My guess is that $a_n \to 0$. So, we could try working from the definition of convergence and show that $a_n \to 0$, but it isn't clear to me how to show that $|a_n| < \epsilon$.

Hints, not complete solutions, are appreciated.

Edit: Could the squeeze theorem potentially be useful here?

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  • $\begingroup$ Did you want $b_k$ on the right? $\endgroup$ – zhw. Jun 15 '16 at 14:54
  • $\begingroup$ @zhw. Yes, thank you. $\endgroup$ – Clarinetist Jun 15 '16 at 14:54
  • $\begingroup$ No, $a_n$ need not approach $0.$ Consider $a_n = 1$ for all $n.$ $\endgroup$ – zhw. Jun 15 '16 at 15:02
  • $\begingroup$ @zhw. Ah, thank you for pointing that out. $\endgroup$ – Clarinetist Jun 15 '16 at 15:03
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Hint: Let $\epsilon>0.$ Then there exists $N$ such that $b_n <\epsilon$ for $n\ge N.$ What happens if $k,l\ge N?$

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  • $\begingroup$ If $k, l \geq N$, then $|x_l - x_k| < b_k < \epsilon$? $\endgroup$ – Clarinetist Jun 15 '16 at 15:17
  • $\begingroup$ Yes, because the smaller of the indices is $\ge N.$ $\endgroup$ – zhw. Jun 15 '16 at 15:45
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Let $\varepsilon>0$. Use the following $|a_{n+m}-a_n|<b_n$ and continue from here. Since $\lim_{n\rightarrow \infty}b_n=0$ then exists $N \in \mathbb{N}$ such that for $n\geq N$ we have $|b_n|<\varepsilon$. Therefore, $|a_{n+m}-a_n|<|b_n|<\varepsilon$.

Edit: As a response to your answer - the squeeze theorem is useful, and you can use if after this hint I gave you.

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  • $\begingroup$ Could the squeeze theorem potentially be useful here? $\endgroup$ – Clarinetist Jun 15 '16 at 14:58
  • $\begingroup$ Yes it could be useful $\endgroup$ – Noy Soffer Jun 15 '16 at 14:59
  • $\begingroup$ I don't see how this answer is helpful. $\endgroup$ – zhw. Jun 15 '16 at 15:08
  • $\begingroup$ @zhw. Nor do I, to be quite frank. $\endgroup$ – Clarinetist Jun 15 '16 at 15:09
  • $\begingroup$ Just edited. Hope this helped. $\endgroup$ – Noy Soffer Jun 15 '16 at 15:18

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