# 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?

• The sequence $a_n = 1$ works. – B. Goddard Mar 21 '18 at 11:34
• $a_n = 1/n$ works also. – Santeri Mar 21 '18 at 11:34

## 2 Answers

$a_n = \cos(\log n))$ 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.

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.

• Why is the hypothesis about bounded-ness redundant? And when apply Cauchy inequality, where is the factor $\sqrt{n-1}$? Am I missing something? – stefano Mar 21 '18 at 13:51
• @stefano: oh, you're right, there should have been a $\sqrt{n-1}$ factor, and the bounded assumption is not redundant. Removing such part since it is not really relevant. – Jack D'Aurizio Mar 21 '18 at 13:55