T\F: if $\lim_{k\to\infty}\sum_{i=n_k}^{n_{k+1}-1}|\alpha_i|= 0$ then $\sum \alpha_k$ converges.

Question

Let $\sum \alpha_k$ be a bounded series of reals.
Suppose $\{n_k\}_{k=1}^{\infty}$ is a strictly-increasing monotonic sequence of naturals s.t $lim_{k\to\infty} (n_{k+1}-n_{k})= \infty$.
Prove or disprove: if $\lim_{k\to\infty}\sum_{i=n_k}^{n_{k+1}-1}|\alpha_i|= 0$ then $\sum \alpha_k$ converges.

I know that if $\{S_k\}_{k=1}^{\infty}$ is the partial sum sequence and we assume that $\{S_{n_k}\}_{k=1}^{\infty}$ converges instead of that it is bounded then it applies. I tried to think about the presented variation and intuitively I guessed it is true.

I can provide a counter example if we omit the absolute values or the limit to infinity of the $n_k$'s differences, however, I struggle to determine the correctness of this argument logically and would like to have a hint or some guidance.

Thanks

Answer 1

There exists a sequence $b_k\to 0$ such that the series $\sum b_k$ is divergent and the partial sums of the series form a bounded sequence. For example $$ b_k={(-1)^l\over 2^l},\quad 2^l\le k<2^{l+1}$$ Let $R_k$ denote the sequence of partial sums of $\sum b_k.$

Let $$a_n=\begin{cases}b_k&n=n_k\\ 0& {\rm otherwise}\end{cases}$$ Then $$S_n=R_k\quad n_k\le n<n_{k+1}$$ $${\rm and}\ \sum_{n=n_k}^{n_{k+1}-1}|a_n|=|b_k|$$ Thus the requirements are satisfied.