real analysis - The limit of the difference of two consecutive sequence members is equal to $0$. Can we conclude that the sequence itself has a limit?

Let $a_n$ be an infinite sequence. The limit of the difference of two consecutive members is equal to $0$. Can we conclude that the sequence itself has a limit?

My attempt:
We have
$$

\lim_{n\rightarrow\infty}{a_n} - \lim_{n\rightarrow\infty}{a_{n-1}} = 0
$$
since as $n$ approaches infinity $a_{n-1}$ gets arbitrarily close to $a_n$ the sequence cannot diverge or be bounded but have no limit.

Is my proof correct and how would I be able to formalize the last sentence? Thanks

Answer

While I can't entirely follow your proof, it seems to be assuming the existence of $\lim_{n\to\infty}a_n$, which is what you're trying to prove or disprove.

Here's an example which shows that you cannot conclude that $\lim_{n\to\infty}a_n$ exists: Let $a_n=1+\frac{1}{2}+\dots+\frac{1}{n}$. Then $a_{n+1}-a_n=\frac{1}{n+1}\to0$, but the sequence $\{a_n\}$ diverges.