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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?

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  • $\begingroup$ You can't, as this is false. $\endgroup$
    – user65203
    Commented Mar 3, 2017 at 17:51

2 Answers 2

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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 $$ a_n + b_n = 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, +1, \ \ 0, \dots$$

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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.

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