1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ The sequence $a_n = 1$ works. $\endgroup$ – B. Goddard Mar 21 '18 at 11:34
  • $\begingroup$ $a_n = 1/n$ works also. $\endgroup$ – Santeri Mar 21 '18 at 11:34
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why is the hypothesis about bounded-ness redundant? And when apply Cauchy inequality, where is the factor $\sqrt{n-1}$? Am I missing something? $\endgroup$ – stefano Mar 21 '18 at 13:51
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Mar 21 '18 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.