Let $\sum a_n$ and $\sum b_n$ converge, $a_n,b_n\geq 0$, does $\sum \min\{a_n,b_n\}$, $\sum \max\{a_n,b_n\}$ converge too? 
Let $\sum\limits_{n=0}^{\infty}a_n$ and $\sum\limits_{n=0}^{\infty}b_n$ be convergent with $a_n,b_n\geq 0$, does $\sum\limits_{n=0}^{\infty}\min\{a_n,b_n\}$ and  $\sum\limits_{n=0}^{\infty}\max\{a_n,b_n\}$ converge too? 

I know, that this was asked here and here in a kind of similar way. The thing is, that in the first link the answer is quite undetailed and the second link contains a different question that is similar but not the same. (That's why I will ask the question yet again.)

My thoughts are, that if both series $\sum_{n=0}^{\infty}a_n,\sum_{n=0}^{\infty}b_n$ converge, that $\sum_{n=0}^{\infty}\min\{a_n,b_n\}$ will pick either value of one of the partial sums and likewise, $\sum_{n=0}^{\infty}\max\{a_n,b_n\}$ will pick a partial sum out of both. With the help of the definition of series, "The series $\sum_{n=0}^{\infty}a_n$ converges, if the partial sum $\sum_{k=0}^{\infty}a_k$ with $k\geq n$ converges", can we apply, that both $\sum_{n=0}^{\infty}\min\{a_n,b_n\}$ and $\sum_{n=0}^{\infty}\max\{a_n,b_n\}$ converge.
Is this enough for a right proof?
 A: Since I don't know what “will pick either value of one of the partial sums” means, I can't tell whether you are right or wrong. But you can do it as follows: since both series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ converge, the series $\sum_{n=0}^\infty(a_n+b_n)$ converges too. And since$$(\forall n\in\mathbb Z_+):\min\{a_n,b_n\},\max\{a_n,b_n\}\leqslant a_n+b_n,$$both series $\sum_{n=0}^\infty\min\{a_n,b_n\}$ and $\sum_{n=0}^\infty\max\{a_n,b_n\}$ converge, by the comparison test.
A: Yes, both converge. Note that both the partial sums are monotone increasing, thus for convergence we only need an upper bound. However, we have
$$ \sum_{j=0}^n \min\{a_j, b_j\} \leq \sum_{j=0}^n \max\{a_j, b_j\} \leq \sum_{j=0}^n (a_j + b_j) \leq \sum_{j\geq 0} a_j + \sum_{j\geq 0} b_j $$
A: 0) $a_n, b_n \ge 0$, resp. series $\sum a_n$, $\sum b_n $ are convergent.
1) $0 \le \min (a_n, b_n) \le a_n$.
By comparison test $\sum \min(a_n, b_n)$ is convergent.
2) $\min (a_n, b_n) +\max (a_n, b_n) =$
$a_n +b_n$;
$0 \le \max(a_n, b_n) =$
$a_n +b_n - \min (a_n, b_n) < a_n + b_n$;
By comparison test $\sum \max(a_n, b_n)$ is convergent since $\sum(a_n +b_n) =\sum a_n +\sum b_n$ is convergent.
A: For $a_n,b_n\ge0$ we have the inequality
$$0\le\min{\{a_n,b_n\}}\le\max{\{a_n,b_n\}}\le a_n+b_n$$
Hence, summing from $n=0$ to $\infty$, gives
$$\sum_{n=0}^\infty0\le\sum_{n=0}^\infty\min{\{a_n,b_n\}}\le\sum_{n=0}^\infty\max{\{a_n,b_n\}}\le \sum_{n=0}^\infty(a_n+b_n)$$
$$0\le\sum_{n=0}^\infty\min{\{a_n,b_n\}}\le\sum_{n=0}^\infty\max{\{a_n,b_n\}}\le \sum_{n=0}^\infty a_n+\sum_{n=0}^\infty b_n$$
by the comparison test.
A: We have that 
$$\sum \min\{a_n,b_n\}\le \sum a_n$$
therefore the series $\sum \min\{a_n,b_n\}$ converges and since
$$\sum \max\{a_n,b_n\}=\sum \frac{|a_n-b_n|+a_n+b_n}{2}=\frac12\sum |a_n-b_n|+\frac12\sum a_n+\frac12\sum b_n $$
also $\sum \max\{a_n,b_n\}$ converges too.
