Finding a Sequence Decreasing faster than a countable collection of sequences

Question

For each $n\in \mathbb N$, let $\alpha_m^{(n)}$ be a sequence in $m$ which strictly decreasing to $0$. Can we find a sequence $\alpha_m$ which also decreases strictly to $0$, such that $\alpha_m=O_n(\alpha_m^{(n)})$ for all $n$? Note the implied constant here may depend on $n$.

For example, if $\alpha_m^{(n)}=m^{-1/n}$, then we can take $\alpha_m=2^{-m}$. But I would like to know if this result holds in general.

Answer 1

How about $\alpha_m:=\min\{\alpha_m^{(n)}:1\leq n\leq m\}$?

Answer 2

What if $\alpha^{(n)}_m = \frac{1}{n+m}$. Then each $\alpha_m$ would have to be 0.