# Proof verification regarding whether a certain property of a sequence implies that it is Cauchy.

I did a proof a while ago that I thought was pretty solid, but I just read something on here that made me doubt that the proposition was even true. The statement I thought I proved was, "Suppose a sequence $\{a_n\}$ has the property that for every $\epsilon>0$, there is an $N$ such that $$\lvert a_{n+1}−a_n\rvert<\epsilon \ \text{whenever} \ n>N.$$ Then $\{a_n\}$ is necessarily Cauchy." However, this seems to be equivalent to, "If $\lvert a_{n+1}-a_n\rvert$ converges to zero, then $\{a_n\}$ is Cauchy," which is evidently false, with the counterexamples of $\{\ln n\}$ and the partial sums of the harmonic series. I, however, cannot see where my proof goes wrong. So could someone please point out the error in either my proof or my reasoning that these two prepositions are equivalent?

$Proof.$ We fix $\epsilon>0$ and $k\in\mathbb N$. Then there is an $N$ such that $$\lvert a_{n+1}-a_n\rvert<\frac{\epsilon}{k} \ \text{whenever} \ n>N.$$ Let $m=n+k$. Then for all $n>N$ \begin{align} \lvert a_m-a_n\rvert&=\left\lvert\sum_{i=1}^k(a_{n+i}-a_{n+i-1})\right\rvert \\ &\le\sum_{i=1}^k\lvert a_{n+i}-a_{n+i-1}\rvert \\ &<\sum_{i=1}^k\frac{\epsilon}{k} \\ &=\epsilon. \end{align} Since this is true independently of our choice of $\epsilon$ or $k$, it is true for all $\epsilon>0$ and for all $n,m>N$.

Hence, $\{a_n\}$ is Cauchy.

The error is that your $N$ depends on $k$.
In order for $(a_n)$ to be a Cauchy sequence, you must show that there is one $N$ such that $$|a_{m}-a_n|<\epsilon$$ for all $m>n$. But if your $N$ depends on $k$, it also depends on $m$, which is not allowed.