How do you prove the sum of monotone sequences is also monotone?

Question

How do you prove the sum of two monotone sequences is also monotone?

Here is my thought process:

Let $a_n$ and $b_n$ be two monotone increasing sequences. Then $\forall n \in N$, $a_n \leq a_{n+1}$ and $b_n \leq b_{n+1}$. Adding both inequalities you get $a_n + b_n \leq a_{n+1} + b_{n+1}$. Therefore in this specific case of both sequences being monotone increasing I have proven their sum is monotone increasing, and so it is also monotone.

But I'm having trouble figuring out how to apply similar logic for an $a_n$ is monotone increasing but $b_n$ is monotone decreasing. I also tried to do proof by contradiction (like supposing the sum is not monotone), but that also lead to nowhere.

How do I prove the other cases?

Answer 1

The result is not true when $a_n$ is monotone increasing and $b_n$ is monotone decreasing.

For example, $$a_n = 0, +1, +1, +2, +2, +3, +3, +4, +4, +5, +5, \dots$$ $$b_n = 0, \ \ 0, -1, -1, -2, -2, -3, -3, -4, -4, -5, \dots $$ gives $$ a_n + b_n = 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, \dots$$

Answer 2

Consider the sequences defined by :

$$a_n=3n+(-1)^n$$

and

$$b_n=-2n$$

It is readily seen that $(a_n)$ is increasing, since for all $n\in\mathbb{N}$ : $a_{n+1}-a_n=3+2(-1)^{n+1}>0.$

Obviously, $(b_n)$ is decreasing.

And finally $(a_n+b_n)$ is known to be non monotonic.