Example of a sequence such that $(n^2(a_n-a_{n+2} -(a_{n+2} -a_{n+4})))$ is bounded but $(n^2(a_n-a_{n+2}))$ is not bounded

Question

Let $(a_n)_{n \in \mathbb{Z} }$ be a real sequence which converges to $0$ as $|n| \to \infty$. It can be shown that if the sequence $(n^2(a_n-a_{n+2}))_n$ is bounded, then the sequence

$$(n^2(a_n-a_{n+2} -(a_{n+2} -a_{n+4})))_n$$

is bounded. I wonder if the converse result is true. Could you give me any help, please?

Answer 1

Take $a_n=\log n$. Then for all $n$ large enough you get

$$ n^2(a_n-a_{n+2})=-n^2\log\left(1+\frac{2}{n}\right)\approx-n^2\cdot\frac 2n=-2n\ \to -\infty. $$

The sequence $(n^2(a_n-a_{n+2}))_n$ is unbounded. On the other hand

$$ n^2(a_n-a_{n+2}-(a_{n+2}-a_{n+4}))=n^2\log\left(1-\frac{4}{(n+2)^2}\right)\approx-n^2\cdot\frac 4{(n+2)^2}\ \to\ -4 $$

and the sequence $(n^2(a_n-a_{n+2}-(a_{n+2}-a_{n+4})))_n$ is bounded.

Answer 2

Sure, just take $a_n = n$.

Then $a_n-a_{n+2} = -2$, and $n^2(a_n-a_{n+2}) = -2n^2$ is unbounded.

And $(a_n-a_{n+2})-(a_{n+2}-a_{n+4}) = 0$ for every $n$, so even multiplying by $n^2$, the result is still always zero and obviously bounded.