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

Mittens
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optical
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2 Answers2

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

Robert Z
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Even if $a_n,b_n\searrow0$ as $n\rightarrow\infty$, it is possible for $$\sum_n\min(a_n,b_n)<\infty$$ when $\sum_na_n=\sum_nb_n=\infty$.

Here is a way to construct an example:

  • Consider positive sequences $u_n=\frac{1}{n^2}$ and $v_n=\frac{1}{\sqrt{n}}$ so that $\sum_nu_n<\infty$ and $\sum_nv_n=\infty$. By alternating sequences $u_n$ and $v_n$ we define sequences $a_n$ and $b_n$ by letting \begin{align} a_{2n-1}&=v_n,\qquad a_{2n}=u_n\\ b_{2n-1}&=u_n,\qquad b_{2n}=v_n \end{align} The series $\sum_na_n$ and $\sum_nb_n$ diverge. However, since $$\min(a_{2n},b_{2n})=\min\Big(\frac{1}{n^2},\frac{1}{\sqrt{n}}\Big)=\frac{1}{n^2}$$ and $$\min(a_{2n-1},b_{2n-1})=\min\Big(\frac{1}{\sqrt{n}},\frac{1}{n^2}\Big)=\frac{1}{n^2}$$ we have that $$\sum_n\min(a_n,b_n)=2\sum_n\frac{1}{n^2}<\infty$$

The sequences $a_n$ and $b_n$ are not monotone. We can use them to build a monotone example as follows:

  • Define $n_1=1$, and for $k>1$ $$ n_k=\min\left\{n\in\mathbb{N}: n>\max\Big(\frac{n_{k-1}a_k}{a_{k-1}},\frac{n_{k-1}b_k}{b_{k-1}}\Big)\right\}$$ Then $\frac{a_k}{n_k}<\frac{a_{k-1}}{n_{k-1}}$ and $\frac{b_k}{n_k}<\frac{b_{k-1}}{n_{k-1}}$. Let $N_0=0$, $N_k=\sum^k_{j=1}n_j$ for $k\geq 1$. Define sequences $\alpha$ and $\beta$ as \begin{align} \alpha_m&=\frac{a_k}{n_k}\qquad\text{if}\quad N_{k-1}+1\leq m\leq N_k\\ \beta_m&=\frac{b_k}{n_k}\qquad\text{if} \quad N_{k-1}+1\leq m\leq N_k \end{align} $\alpha_m,\beta_m\searrow0$ as $m\rightarrow\infty$ and
    \begin{align} \sum_m\alpha_m&=\sum^\infty_{k=1}\sum^{N_k}_{j=N_{k-1}+1}\frac{a_k}{n_k}=\sum_ka_k=\infty\\ \sum_m\beta_m&=\sum^\infty_{k=1}\sum^{N_k}_{j=N_{k-1}+1}\frac{b_k}{n_k}=\sum_kb_k=\infty \end{align} and $$\sum_m\min(\alpha_m,\beta_m)=\sum_k\min(a_k,b_k)<\infty$$
Mittens
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