Serie of max of two values diverges implies that the max of the series diverges.

Question

Context :

In the context of simplifying the conditions presented in this post, the following question arise.

Question :

Suppose that $0<a_k,b_k$ for $k\in\mathbb N$ are such that $\sum_{k\geq 1} \max(a_k,b_k)=\infty$, then $\max\left( \sum_{k\geq 1} a_k, \sum_{k\geq 1} b_k \right)=\infty$.

If this is true, then it is actually an if and only if, indeed if $\sum_{k\geq 1} a_k=\infty$ then of course $\sum_{k\geq 1} \max(a_k,b_k)\geq \sum_{k\geq 1} a_k=\infty$.

Intuition :

My intuition is that it should be true, indeed if $\max\left( \sum_{k\geq 1} a_k, \sum_{k\geq 1} b_k \right)<\infty$, then $a_k,b_k$ are both $o(1/k)$ and so $\max(a_k,b_k)$ also is $o(1/k)$ and I feel like this would imply that $\sum_{k\geq 1} \max(a_k,b_k)$. Of course this is just intuition and it might be wrong, I do not know any result of the form relating complexity of the coefficients and convergence of series.

Answer 1

The intuition is wrong because being $o(1/k)$ does not imply the convergence of the series. Take for instance $a_k=\frac{1}{k \log k}$ for $k \geq 2$. In this case $a_k=o(\frac{1}{k})$ but $\sum_{k\geq 2} a_k=\infty$.

However, being $0<a_k,b_k$, the thesis is still true because: $$\infty \ = \ \sum_{k\geq 1} \max(a_k,b_k) \leq \ \sum_{k\geq 1} (a_k+b_k) \ = \ \sum_{k\geq 1} a_k + \sum_{k\geq 1} b_k \ ,$$

which implies that at least one of the two terms on the right-hand side is $\infty$, i.e. $\max \left( \sum_{k\geq 1} a_k \ , \sum_{k\geq 1} b_k \right) = \infty$

Answer 2

Series with non-negative terms are convergent if (and only if) their partial sums are bounded above.

If both $\sum_k a_k$ and $\sum_k b_k$ are convergent then so is $\sum_k \max(a_k, b_k)$, because $$ \max(a_k, b_k) \le a_k + b_k \, . $$ Therefore, if $\sum_k \max(a_k, b_k)$ is divergent then at least one of $\sum_k a_k$ and $\sum_k b_k$ must be divergent.