Example of a conditionally convergent series $\sum_{n=1}^\infty b_n$ such that $n^2(b_n-b_{n+1})$ is bounded

Question

Let $(b_n)_{n \in \mathbb{N}}$ be a real sequence such that $(nb_n)$ is bounded. I know that if the series $\sum_{n=1}^\infty b_n$ is conditionally convergent, then $(n^2b_n)_n$ is not bounded. But, I wonder if there is an example of a real sequence $(b_n)_n$ such that the corresponding series is conditionally convergent and the sequence $n^2(b_n−b_{n+1})_n$ is bounded.

For now, I know that $(b_n)$ cannot be an alternating sequence. I also know that the series associated to the sequences $(\sin(\sqrt{n})/n)_{n \in \mathbb{N}}$ and $((-1)^n/(n \ln(n)))_{n \in \mathbb{N}}$ are conditionally convergent, but in these cases the sequence $n^2(b_n−b_{n+1})_n$ is unbounded.

I have also tried to show that if $n^2(b_n−b_{n+1})_n$ is bounded, then the series $\sum_{n=1}^\infty b_n$ must be absolutely convergent, but I don't have a proof yet.

Answer 1

The answer is yes, such an example exists.

For instance, let $b_1=0$ and $b_n=(-1)^j h_j(n)$ for $j=1,2,\dots$ and $n\in N_j:=\{2^j,\dots,2^{j+1}-1\}$, where $$h_j(n):=c_j\frac{\min(n-2^j,2^{j+1}-n)}{2^{j-1}},\quad c_j:=\frac1{j 2^j}.$$

Then for $n\in N_j$ we have $$n^2|b_{n+1}-b_n|=O\Big((2^j)^2c_j\frac1{2^{j-1}}\Big) =O\Big(\frac1j\Big)=O(1).$$ Next, for $$s_m:=\sum_{n=1}^m b_n\quad\text{and}\quad t_j:=\sum_{n\in N_j}b_n$$ we have $$t_j=\frac1{2j}$$ and hence for $k=1,2,\dots$ $$s_{2^k}=\sum_{j=1}^k (-1)^j t_j =\frac12\sum_{j=1}^k \frac{(-1)^j}j,$$ so that $s_{2^k}\to-\frac{\ln 2}2$ as $k\to\infty$. Also, for $m\in N_j$ the value of $s_m$ is between $s_{2^k}$ and $s_{2^{k+1}}$. So, $s_m\to-\frac{\ln 2}2$ as $m\to\infty$.

Also, $\sum_{n=1}^\infty|b_n|= \sum_{j=1}^\infty t_j=\sum_{j=1}^\infty \frac1{2j}=\infty$. So, the series $\sum_{n=1}^\infty b_n$ is (only) conditionally convergent. $\quad\Box$