# Termwise max (resp. min) of two convergent (resp. divergent) series

If $\sum_{n = 0}^\infty a_n$ and $\sum_{n = 0}^\infty b_n$ converge, does $\sum_{n = 0}^\infty \max\{a_n, b_n\}$ also converge?

Similarly, if $\sum_{n = 0}^\infty a_n$ and $\sum_{n = 0}^\infty b_n$ diverge, does $\sum_{n = 0}^\infty \min\{a_n, b_n\}$ also diverge?

My intuition tells me the answer to both of these questions is yes, but I don't know how to verify it.

Edit: $a_n \ge 0, b_n \ge 0\ \forall n$

## 1 Answer

Hint. As regards the max case, note that $$0\leq\max\{a_n, b_n\}\leq a_n+b_n$$ For the min case take $$a_n=1+(-1)^n+e^{-n}>0\quad\mbox{and}\quad b_n=1+(-1)^{n+1}+e^{-n}>0 \implies \min\{a_n, b_n\}=e^{-n}.$$ What may we conclude?

P.S. As regards the max case, without the assumption of non negativity, we have a counterexample: $$a_n=\frac{(-1)^n}{n+1}\quad\mbox{and}\quad b_n=\frac{(-1)^{n+1}}{n+1}\implies \max\{a_n, b_n\}=\frac{1}{n+1}.$$

• I initially forgot to say it in my question, but the terms are nonnegative. – OpticAl Sep 19 '17 at 16:51
• @OpticAl I edited my answer. – Robert Z Sep 19 '17 at 16:57
• For the max: This is so simple, why haven't I thought of that? For the min, this is also so simple, but I thought of something even simpler: $a_n$ is 1 if $n$ is odd and 0 if $n$ is even, $b_n = 1 - a_n$. – OpticAl Sep 19 '17 at 17:23
• @OpticAl Yes, well done! You are correct! – Robert Z Sep 19 '17 at 18:06