If $\sum_{n=1}^\infty a_n$and $\sum_{n=1}^\infty b_n$ are convergent and $a_n, b_n > 0$ then so is $\sum_{n=1}^\infty \max\lbrace a_n, b_n\rbrace$

Question

The question given is that in the title;

If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are convergent and $a_n, b_n > 0$ then so is $\sum_{n=1}^\infty \max\lbrace a_n, b_n\rbrace.$

The answer given in my textbook is as follows;

Certainly $\sum_{n=1}^\infty(a_n + b_n)$ is convergent. Since $\max\lbrace a_n, b_n\rbrace < a_n + b_n$, it follows from the comparison test that $\sum_{n=1}^\infty \max\lbrace a_n, b_n\rbrace$ is convergent.

I understand this solution but I wrote something along the lines of

Since both series are convergent we can simply say that $$A = \sum_{n=1}^\infty a_n\ \ \text{and} \ \ B = \sum_{n=1}^\infty b_n.$$ Then $\max\lbrace a_n, b_n \rbrace = \frac{1}{2}(a_n + b_n + \lvert a_n - b_n\rvert)$ and so \begin{align} \sum_{n=1}^\infty \max\lbrace a_n, b_n\rbrace &= \frac{1}{2}\sum_{n=1}^\infty a_n + b_n + \lvert a_n - b_n\rvert\\ &= \frac{1}{2}(A + B + \lvert A - B\rvert) \end{align}

Is this valid too?

Answer 1

It won't be true in general that $\sum_{n=1}^{\infty}|a_n-b_n|=|A-B|$. For instance, suppose that $a_n$ is the sequence $$ 1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\dots $$ and $b_n$ is the sequence $$ \frac{1}{4},1,\frac{1}{16},\frac{1}{9},\dots$$

Then $A=B$ so $|A-B|=0$, but $\sum_{n=1}^{\infty}|a_n-b_n|>0$.

It is true that $\sum_{n=1}^{\infty}|a_n-b_n|$ converges, but the easiest proof is to note that $|a_n-b_n|\leq a_n+b_n$, so this gets you right back to $A+B$ anyway.