# Sum $\sum c_n$ That's Unbounded iff The Sums $\sum a_n$ and $\sum b_n$ Are

Suppose we are given two arbitrary sums in $$\mathbb{R}$$, $$A_n=\sum_{i=1}^n a_i$$ and $$B_n=\sum_{i=1}^n b_i$$. Is there any intuitive sum $$C_n=\sum_{i=1}^n c_i$$ based on $$A_n$$ and $$B_n$$ such that $$C_n$$ is unbounded if and only if $$A_n$$ and $$B_n$$ are both unbounded?

Obvious considerations such as adding, multiplying, or dividing the sequences/sums clearly don't work. Really, $$C_n$$ could just be a sequences rather than a series for my needs. For example something like $$C_n=\left\|\sum_{i=1}^n \langle a_i,b_i\rangle\right\|_2$$ (which clearly doesn't work).

My goal is to simplify a theorem that says some event occurs iff these two series are unbounded, and I would like to simply the statement to a single condition one can assess. Certain methods may achieve this but I'm looking for answers that would be a "good" condensation of the criteria, not one that feels like a convoluted encoding of no independent merit.

• Could we have restrictions such as $a_i, b_i > 0$? Commented Mar 1, 2019 at 17:04
• @TomChen It could still be useful to see, but in my case $a_i,b_i\in\{1, -1\}$. Commented Mar 1, 2019 at 17:05
• If $a_n,b_n$ are non-negative sequences, then $c_n = \min\{a_n,b_n\}$ does the trick Commented Mar 1, 2019 at 17:05
• is $c_n = \mathbf 1_{\sup_k |A_k| > n \text{ and } \sup_k |B_k| > n }$ cheating? Commented Mar 1, 2019 at 17:12
• @CalvinKhor I wouldn't say cheating, but at that point why not just confirm that both sequences are unbounded since you have to compute each individually to compute $c_n$. Commented Mar 1, 2019 at 17:15

How about $$C_n = [\exp(-|A_n|) + \exp(-|B_n|)]^{-1}$$? Of course, this would imply that $$c_k$$ is a function of $$\{a_1, \cdots, a_k\}$$ and $$\{b_1, \cdots, b_k\}$$.

• I'll accept this if there are no better answers. I'm hoping for an answer that is less computation than simply computing both sums. Because in this one could just as easily check that $A_n$ and $B_n$ are unbounded since they are necessary to compute individually when computing $C_n$. Commented Mar 1, 2019 at 17:13