# How to prove non-monotonic sequence

I am asked to prove a sequence is not monotone, the sequence is increasing, but not every term is greater than the preceding so I know by definition the sequence is not monotone. I was wondering what technique I can use to prove a sequence is not monotone because simply showing that the terms are not increasing monotonously is not going to cut it as a proof. Also my sequence is in Z+ and all the values I get are positive, (just not increasing monotonously). Any help would be greatly appreciated, thanks in advance

• FInding a single counterexample to monotonicity is sufficient. Oct 7 '17 at 20:40
• Why would "showing that the terms are not increasing monotonously" not work? All you have to do is find some positive integers $n,m$ such that $n<m$ and $x_n>x_m$. Oct 7 '17 at 20:40
• show us your sequence that would it make simpler to talk about it Oct 7 '17 at 20:55
• @JohnGriffin that is wrong. think it could be monotone increasing or decreasing Oct 7 '17 at 20:57
• The sequence is (x+((-1)^x)2). This is what I have using the above answers. Since the sequence is increasing we would assume if it were monotonic we would find all Sn < Sm whenever n < m. But since we can find an m < n when Sn < Sm we know the sequence is not monotonic. In particular we choose n=2 and m=3, then Sn=4, Sm=1 thus 2 < 3 (Sn < Sm) but 1 < 4 (m < n). I'm sure my teacher will say this is not a formal proof.
– jack
Oct 7 '17 at 21:05

$s_n$ is monotonically increasing means

$$\forall n \in \mathbb{N}:\; s_n<s_{n+1}.\tag 1$$ Check if the sequence $$s_n=n+2(-1)^n$$ is monotonically increasing.

Proof:

We have

$$s_2=4>1=s_3.$$ Therefore the sequence is not monotonically increasing, because this contracdicts $(1)$ for $n=2$.

$\blacksquare$

That is the proof. What are you missing?

A more formal way:

We have $$4>1\\ \implies s_2>s_3 \\ \implies \lnot (s_2<s_3) \\ \implies\exists n \in \mathbb{N}:\lnot (\;s_n<s_{n+1})\\ \iff \lnot \forall n \in \mathbb{N}:\;s_n <s_{n-+1}$$

which is the negation of $(1)$