We still know that: if a sequence of real numbers converges, it is a bounded sequence and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.
I need an example, which is a bounded sequence of real numbers $\left(a_n \right)_{n\in\mathbb N^*}$ that do not converge and $ \lim_{n \to \infty} \left (a_ {n + 1} - a_n \right) = 0 $.
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Take, for instance, the sequence$$0,1,\frac12,0,\frac13,\frac23,1,\frac34,\frac24,\frac14,0,\frac15,\ldots\tag1$$For any $N\in\Bbb N$, you have $|a_{n+1}-a_n|<\frac1N$ of $n$ is large enough, which means that $\lim_{n\to\infty}(a_{n+1}-a_n)=0$. However, the sequence doesn't converge. Actually, every element of $[0,1]$ is the limit of some subsequence of $(1)$.
Take any unbounded sequence $a_n$ with differences that go to $0$ (surely you know some such sequences). Consider the sequence $b_n=a_{n+1}-a_n$. This gives $a_n+b_n=a_{n+1}$. Now instead consider the sequence $c_n$ given by $c_0=a_0$ and $$ c_{n+1}=c_n\pm b_n $$ where the sign is strategically chosen so that the sequence doesn't converge, but is bounded. Say, for instance, you choose signs so it just grows past $1$, then decreases until it just decreases past $-1$, then up again to $1$, and so on.
You need to show that $c_n$ will keep going up and down indefinitely, and the non-convergence of $a_n$ is crucial here. You also need to show that it is bounded, and $b_n\to 0$, or rather the boundedness of $b_n$, is crucial here.