1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $q<2.$) $\endgroup$ – DanielWainfleet Nov 3 '19 at 14:11
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

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.