# 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?

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.

• We could also use $\min \{a_n,b_n\}\le a_n$ for the $\min$ series.... An application: For $2\le p<\infty$ and $1/p+1/q=1,$ if $x=(x_n)_n\in l^p$ and $y=(y_n)_n\in l^q$ then $\sum_n x_ny_n$ converges. Proof: Let $z_n=0$ if $y_n=0.$ If $y_n\ne 0$ let $y_n=z_n|z_n|^{q-2}.$ Then $(z_n)_n\in l^p$ and $|x_ny_n|\le \max \{|x_n|^p, |z_n|^p\}.$ (I distinguish the case $y_n=0$ in case $q<2.$) – DanielWainfleet Nov 3 '19 at 14:11

Yes, both converge. Note that both the partial sums are monotone increasing, thus for convergence we only need an upper bound. However, we have $$\sum_{j=0}^n \min\{a_j, b_j\} \leq \sum_{j=0}^n \max\{a_j, b_j\} \leq \sum_{j=0}^n (a_j + b_j) \leq \sum_{j\geq 0} a_j + \sum_{j\geq 0} b_j$$

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.

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.

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.