Proving that $a_n$ is not monotonically increasing if $a_n\ge 0$ and $\lim a_n=0$

Had an exam today and there was the following question:

Let $$a_n$$ be a sequence such that for all $$n\in \mathbb N$$ we have $$a_n \geq 0$$ and $$\lim a_n = 0$$. Prove that $$a_n$$ is not monotonically strictly increasing.

It feels like an easy question but I have no idea how to prove it.

• This is false because you can take the constant sequence $0$. I guess you mean it is not monotonically strictly increasing. – Mark Jun 18 at 15:43
• @Mark Yes, thank you. I have edited. – vesii Jun 18 at 15:47

3 Answers

If it is increasing, then, for every $$n\ge 3$$, $$|a_n|>a_2>0$$ which contradicts the condition that $$\lim a_n=0$$.

• why it contradicts? Also why did you use abs? if its positive. – vesii Jun 18 at 15:47
• Do you know the definition of the limit? Take $\epsilon=a_2$. The distance of the next elements in the sequence from $0$ will never be smaller than $\epsilon$. – Mark Jun 18 at 15:49
• @vesii take limits, you get $\lim |a_n|\geq a_2>0$ – Julian Mejia Jun 18 at 15:49
• So you use the Squeeze theorem? if possible to add a bit more explanation. I understand the logic on why its not correct, but the problem is to write it formally. – vesii Jun 18 at 15:51

Let $$(x_n)_{n\ge 0}$$ be a stictly increasing sequence that converges to $$L$$.

Then $$L = \text{sup}(x_n)$$ and for any integer $$k \ge 0$$, $$\;x_k \lt L$$.

So if $$(x_n)_{n\ge 0}$$ is strictly increasing, converges to $$L$$, and all $$x_n \ge 0$$, then $$L \gt 0$$.

As of $$\lim\limits_{n\to\infty}a_n=0$$ we have $$\forall\epsilon>0\exists N(\epsilon):|a_n|<\epsilon$$ $$\forall n\in\mathbb{N}$$ with $$n>N(\epsilon)$$

Now assume $$(a_n)_{n\in\mathbb{N}}$$ is monotonically strictly increasing:

By definition we get $$a_{n+1}=a_n+x_n$$ where $$x_n\in\mathbb{R}$$ and $$x_n>0$$.

Because of $$a_n\geq0$$ $$\forall n\in\mathbb{N}$$ and $$x_n>0$$ $$\forall n\in\mathbb{N}$$ we get $$0.

Therefore $$\forall n\in\mathbb{N}$$ with $$n\geq2$$ we have $$0 or rather $$0\leq a_n-x_1\Leftrightarrow x_1\leq a_n$$.

If we now set $$\epsilon:=\frac{x_1}{2}$$ we get $$|a_n|=a_n>\epsilon$$ $$\forall n\in\mathbb{N}$$ with $$n\geq2$$ which means $$(a_n)_{n\in\mathbb{N}}$$ can not be monotonically strictly increasing as it would hurt the requirement $$\lim\limits_{n\to\infty}a_n=0$$.

I know it's a lot of text but I hope it's helpful :)