Consider the sequence $x_n=\sum_{k=1}^n \frac{1}{k}$, then $\vert x_{n+1}-x_n\vert=\vert\sum_{k=1}^{n+1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\vert=\frac{1}{n+1}$. Here, $\vert x_{n+1}-x_n\vert$ can be made less that $\epsilon\; \forall\epsilon>0$ making $\langle x_n\rangle$ a Cauchy Sequence [Suppose that $\langle x_n\rangle$ is a sequence of real numbers satisfying:$\forall \epsilon>0$ there exist $n_0$ such that $\vert x_{n+1}-x_n\vert<\epsilon\;\forall n\ge n_0$] and hence convergent.
But, $x_n=\sum_{k=1}^n \frac{1}{k}\implies \lim x_n=\sum_{k=1}^{\infty }\frac{1}{k} $ is divergent making <$x_n$> a divergent sequence.
So, both the above arguments are true which cannot be the case...please point out the mistake I'm making.
Your first argument is not correct.
A sequence $(x_n)_n$, the difference of whose consecutive terms is "eventually small", need not be convergent.
The sequence $( \sum_{k=1}^{n} 1/k )_n$ is an example. It diverges.