I'm currently studying real analysis with the book of "Understanding Analysis" by Abott, I was making Ex. 2.6.5, which states:
"Consider the following (invented) definition: A sequence $(s_n)$
is pseudo-Cauchy if, for all $\epsilon$ > 0, there exists an $N$ such that if $n ≥ N$, then
$|s_{n+1} − s_n| < \epsilon$. Decide which one of the following two propositions is actually true. Supply
a proof for the valid statement and a counterexample for the other.
(i) Pseudo-Cauchy sequences are bounded.
(ii) If $(x_n)$ and $(y_n)$ are pseudo-Cauchy, then $(x_n + y_n)$ is pseudo-Cauchy as
well."
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So it appears that (i) is false, but I "managed" to provided a proof for it, so I would like to know where I went wrong, here's what I did:
Let $(s_n)$ be pseudo-Cauchy, therefore there exist $N$ such that if $n≥N$, then $|s_{n+1}-s_{n}| <\frac{\epsilon}{k}$, for some $k\in \mathbb{N}$. Without loss of generality assume $m>n$ and let $m =n+k$. Suppose that $m,n≥N$, then $$ |s_m-s_n|=|s_{n+k}-s_n|=|s_{n+k}-s_{n+k-1}+s_{n+k-1}+\dots+s_{n+1}-s_n|\\\leq|s_{n+k}-s_{n+k-1}|+\dots+|s_{n+1}-s_n|<\frac{\epsilon}{k}+\dots+\frac{\epsilon}{k}=\epsilon$$That is $|s_m-s_n|<\epsilon$, therefore $(s_n)$ is Cauchy and is bounded.
Any insight would be appreciate it, thanks.
Best Answer
Yes, the first assertion is indeed false (take $s_n=\log(n)$, for instance). The problem with your proof lies in that $k$. What you actually proved was that, for each $k\in\Bbb N$ and for each $\varepsilon>0$, $|s_{n+k}-s_n|<\varepsilon$, if $n$ is large enough. So what? It doesn't follow from that that $(s_n)_{n\in\Bbb N}$ is a Cauchy sequence. You did not prove that you have $|s_m-s_n|<\varepsilon$ if $m$ and $n$ are large enough.