Here is an interesting problem:
assume a square summable positive sequence $\{a_k\}$
$$\sum_{k=1}^{K} a_k^2 < +\infty, (\text{$K$ is any positive integer})$$
If this sequence is not summable, i.e., $\sum_{k=1}^{K} a_k \rightarrow +\infty$, then given a window of width $s$, does there exists a constant $D$, s.t.
$$\forall K\ge s, \sum_{k=K-s}^{K-1} a_k \le D*a_K ~~~?$$
I guess $D$ exists, and $D$ eventually converge to $s$ as $K \rightarrow +\infty$. But I m not sure, can anyone give a rigorous proof or a counter example ?
[Math] square summable and summable
sequences-and-series
Best Answer
I think this is wrong. Consider the sequence $$a_k:=\cases{2^{-k}&$\quad(k$ even)\cr {1\over k}&$\quad(k$ odd)$\ $.\cr}$$ This sequence is square summable, but the $a_K$ with even $K$ are just too small to allow for a universal $D$, given $s\geq2$.