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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.

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    $\begingroup$ This is false because you can take the constant sequence $0$. I guess you mean it is not monotonically strictly increasing. $\endgroup$ – Mark Jun 18 at 15:43
  • $\begingroup$ @Mark Yes, thank you. I have edited. $\endgroup$ – vesii Jun 18 at 15:47
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If it is increasing, then, for every $n\ge 3$, $$|a_n|>a_2>0$$ which contradicts the condition that $\lim a_n=0$.

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  • $\begingroup$ why it contradicts? Also why did you use abs? if its positive. $\endgroup$ – vesii Jun 18 at 15:47
  • $\begingroup$ 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 $\epsilon$. $\endgroup$ – Mark Jun 18 at 15:49
  • $\begingroup$ @vesii take limits, you get $\lim |a_n|\geq a_2>0$ $\endgroup$ – Julian Mejia Jun 18 at 15:49
  • $\begingroup$ 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. $\endgroup$ – vesii Jun 18 at 15:51
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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 $L \gt 0$.

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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 $0<a_1+x_1=a_2$.

Therefore $\forall n\in\mathbb{N}$ with $n\geq2$ we have $0<a_n$ 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 :)

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