# Cauchy Sequences, not converging to zero

True or False? If $\{x_n\}$ and $\{y_n\}$ are Cauchy and $x_n + y_n > 0$, for all $n\in\mathbb{N}$, then $\left\{\frac{1}{(x_n + y_n)}\right\}$ cannot converge to zero.

I believe the claim to be False: If both sequences are Cauchy, then they are convergent and therefore bounded. Then there exists a positive number for which the sequences is less than or equal to that positive number. Then $1/(x_n + y_n) \leq M$ for all $n$. However, by Cauchy Theorem, a sequence must approach a real value. Then in this case it would be zero, but $x_n + y_n > 0$,therefore it cannot approach zero.

Does this make sense? Can someone please help me, clarify? Any feedback would be really appreciated. Thank you

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To disprove a general statement, you should come up with a counterexample. All you would need in this case is to find one satisfactory set of Cauchy sequences where the statement does not apply. However, as the answer points out, that's impossible. –  Omnomnomnom Sep 20 '14 at 5:07
In fact, your statements all seem to be proving the statement true, so your answer would have been correct if you simply wrote "true" rather than "false" and kept the same explanation –  Omnomnomnom Sep 20 '14 at 5:08

It is true. If the sequences $\{x_n\}$ and $\{y_n\}$ are Cauchy, then they are both bounded. Thus there is an $M\gt 0$ such that $|x_n|\lt M$ for all $n$, and $|y_n|\lt M$ for all $n$. It follows that $|x_n+y_n|\lt 2M$ for all $n$. Thus the sequence $\{\frac{1}{x_n+y_n}\}$ cannot have limit $0$.
Remark: We did not need the condition $x_n+y_n\gt 0$. We have left a few little gaps in the argument, which for a fully detailed proof should be filled. We assumed that you already had seen a proof of the fact that Cauchy sequences are bounded. The fact that if $|x_n+y_n|\lt 2M$, then the reciprocal sequence cannot have limit $0$ can, if you wish, be proved by a straightforward $\epsilon$-$N$ argument.