Is it possible for $x_{n+1}-x_{n}$ to converge to $0$, but the limit of $x_n$ to not exist? I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series). But then I thought "is it possible for the limit of $x_n$ to not exist at all?" and I think the answer is no,because I couldn't find any examples, but I am not sure. Could you please tell me what your opinion is on this?
Thanks in advance!

The answer is yes!
Not only you can find examples were the limit is not finite but is for example $\infty$. You can also have examples were a full interval is a set of limit points of the sequence.
Look at a sequence that will slide back and forth from $0$ to $1$ with steps decreasing to zero for example. Let's for example consider $\{x_n\}$ whose initial terms are
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \dots$$
Formal definition of $\{x_n\}$ is
$$
x_n=\begin{cases}
\frac{1}{2} &\text{ for } n= 1\\
\frac{n}{k+2} - \frac{k(k+1)}{2(k+2)} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2}
\end{cases}$$
The values of the sequence $\{x_n\}$ are in $(0,1) \cap \mathbb Q$. Moreover, one can notice that $\{x_n\}$ takes each rational number of $(0,1)$ as value an infinite number of times. Indeed for $\frac{p}{q} \in (0,1)$ with $1 \le p \lt q$ and $m \ge 1$ we have $$
\frac{(mq-2)(mq-1)}{2} \lt \frac{(mq-2)(mq-1)}{2} + mp$$ and $$\begin{aligned}
\frac{(mq-2)(mq-1)}{2} + mp &\le \frac{(mq-2)(mq-1)}{2} + m(q-1)\\
&\le \frac{(mq-2)(mq-1)}{2} + mq-1\\
&= \frac{(mq-1)mq}{2}
\end{aligned}$$ Hence $$\begin{aligned}
r_{\frac{(mq-2)(mq-1)}{2} + mp} &= \frac{(mq-2)(mq-1)}{2mq} + \frac{mp}{mq} - \frac{(mq-2)(mq-1)}{2mq}\\
&= \frac{mp}{mq}\\
&= \frac{p}{q}
\end{aligned}$$ proving the desired result. As the rational numbers of the segment $(0,1)$ are dense in $[0,1]$, we can conclude that the set of limit points of $\{x_n\}$ is exactly the interval $[0,1]$.
See here for additional examples.
