Counterexample for series I am searching for an example of bounded and divergent sequence $\left(  a_{n}\right)
_{n\in\mathbb{N}}$ such that the series
$$
\sum_{n\geq1}\left\vert a_{n+1}-a_{n}\right\vert ^{2}
$$
be convergent. It is easy to see that the sequence
$$
a_{n}=\sum_{k=1}^{n}\dfrac{1}{k}%
$$
is a good example such that the series be convergent, but it is not bounded.
Also I tried with the bounded sequence
$$
a_{n}=\cos\sqrt{\pi^{2}n}%
$$
for $n\in\mathbb{N}$, but the series seems to be divergent. How can I find
good example? Does an such example exists?
 A: $a_n = \cos(\log n))$ should work.  
Now
$$
\log(n+1)-\log(n) =\log\left(1+\frac{1}{n}\right)\le \frac{1}{n}
$$
Using the mean value theorem on $\cos$, and the fact that $|\sin x| \le 1$
$$
|\cos(\log(n+1))-\cos(\log(n))| \le \frac{1}{n}
$$
So $|a_{n+1}-a_n|^2 \le \frac{1}{n^2}$, and by comparison the series $\sum_{n=1}^\infty |a_{n+1}-a_n|^2$ converges.
Next: $\log(n) \to \infty$; for an even integer $m$, when $\log(n) < \pi m < \log(n+1)$ we have $|a_n - 1| < \frac{1}{n}$.  For an odd $m$ we have $|a_n+1|<\frac{1}{n}$.  This tells us $\limsup a_n = 1$ and $\liminf a_n = -1$.  The sequence $a_n$ is bounded and divergent.
A: I always believed that bounded and divergent is an oxymoron, so I will stick to the terminology I am used to: we want to find a bounded sequence $\{a_n\}_{n\geq 1}$ such that $\lim_{n\to +\infty}a_n$ does not exist but $\sum_{n\geq 1}b_n^2$ is convergent, where $b_n=a_{n+1}-a_n$.  
Does $\color{red}{a_n=\sin(\log n)}$ work? Let us see.
Clearly $|a_n|\leq 1$ and $\lim_{n\to +\infty}a_n$ does not exist, and due to Lagrange's theorem
$$ \left|\sin(\log(n+1))-\sin(\log n)\right|=\left|\frac{\cos\log(n+\xi)}{n+\xi}\right|=O\left(\frac{1}{n}\right)\qquad (\xi\in(0,1))$$
so $\sum_{n\geq 1}b_n^2$ is convergent. It works.
