unbounded sequence convergent subsequence An unbounded sequence $a_n$ can have a convergent subsequence for example $a_n = n$ if $n$ is odd and $a_n = 1/n$ if $n$ is even. What if it is unbounded and $|a_{n+1} - a_n| \to 0$? Can I conclude that it has no convergent subsequences?

Nope! We can turn the harmonic series into a counter-example by taking $H_n = \sum_{k=1}^n 1/k$ and definine $a_n = H_n \sin (2\pi H_n)$.
To see that this has a subsequence that converges to $0$, we'll see that $\sin(2\pi H_n)$ gets extremely close to $0$ infinitely often.
Specifically, let $M \in \mathbf{Z}_{>0}$ and let $k$ be such that $H_{k-1} < M \leq H_k$.
Now,it is a standard fact about the harmonic series that $H_k \leq \log k$, so that $k \geq e^{H_k} \geq e^M$ and so
\begin{align*}
|H_k - M|
\leq H_k - H_{k-1}
= \frac{1}{k}
\leq e^{-M}.
\end{align*}
This means that, for every $M$, there is a $k$ such that
$$
|H_k \sin(2\pi H_k)|
= |H_k \sin(2\pi(H_k - M))|
\leq 2\pi H_k |H_k - M|
\leq 2\pi (M+1) e^{-M},
$$
and so, along this subsequence, $a_k \rightarrow 0$.
We can do a similar analysis to show that $H_n$ also gets very close to $1 + \frac{1}{2}\mathbf{Z}_{>0}$, where $a_n \approx H_n$ and so $a_n$ diverges along this subsequence.
