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Take $\boxed{a_n=(-1{)}^n}$. Then $|a_n/n|=1/n\to 0$, but $a_{n+1}-a_n=-2(-1{)}^n$ alternates between $2$ and $-2$. Thus convergence of the scaled sequence does not imply convergence of its successive differences.

A sequence converges to $\ell$ precisely whenFor a finite collection of convergent sequences, choose $N_j$ for the same tolerance $\epsilon$ in row $j$, and let $N={\max }_{1\le j\le k}{N}_j$. If $n\ge N$, every available $y_n^{(j)}$ lies within $\epsilon$ of $\ell$, and so does any selected $x_n$. Every such finite rowwise selection converges to $\ell$. The finiteness is what permits a single maximum of the cutoffs.

For the infinite collection, defineFor every fixed $j$, this sequence is eventually exactly $\ell$, and hence converges to $\ell$. But its diagonal selection is $x_n=y_n^{(n)}=\ell +(-1{)}^n$, which has two distinct subsequential limits and does not converge. A moving row index can always choose the exceptional entry before that row reaches its eventual behavior.