1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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: $$a_n=\frac{(-1)^n}{n+1}\quad\mbox{and}\quad b_n=\frac{(-1)^{n+1}}{n+1}\implies \max\{a_n, b_n\}=\frac{1}{n+1}.$$

$\endgroup$
  • $\begingroup$ I initially forgot to say it in my question, but the terms are nonnegative. $\endgroup$ – OpticAl Sep 19 '17 at 16:51
  • $\begingroup$ @OpticAl I edited my answer. $\endgroup$ – Robert Z Sep 19 '17 at 16:57
  • $\begingroup$ 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, $b_n = 1 - a_n$. $\endgroup$ – OpticAl Sep 19 '17 at 17:23
  • $\begingroup$ @OpticAl Yes, well done! You are correct! $\endgroup$ – Robert Z Sep 19 '17 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.