# Proof of the divergence of a monotonically increasing sequence

Say that the limit of sequence $(A_n)$ as $n \to \infty$ equals $+\infty$ if for every $r \in \mathbb R$, there is an integer $N$ such that $(A_n) > R$ for all $n \geq N$.

Show that a divergent monotone increasing sequence converges to $+\infty$ in this sense.

I am having trouble understanding how to incorporate in my proof the fact that the sequence is monotonically increasing.

Any help would be appreciated,

Thanks

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What is required for a monotone increasing sequence to converge?

By$\; (*):\;$ Since $A_n$ is a monotone increasing sequence, if $A_n$ were bounded above, then it would converge to some value $L < +\infty$. But we are given $A_n$ is a divergent monotone increasing sequence. Hence $A_n$ cannot be bounded above; i.e., $A_n$ has no upper bound. (That's simply applying the contrapositive of the monotone convergence theorem).

That is, there is no $M > 0$ such that $(A_n)$ is bounded above by $M$. This means $$\forall M>0,\; \exists N\in \mathbb{N}:A_N>M.$$

And since we are given that $(A_n)$ is monotone increasing, $n\ge N\implies A_n>A_N>M$.

This holds for $M'<M$ too, so $$\forall M>0,\;\;\exists N\in \mathbb{N}:\ n\ge N\implies A_n>M$$ and hence, $A_n\to +\infty$.

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 Why $\exists M$ and not $\forall M$? I'd have bet it should've been $\forall M$ but since you are two to use $\exists M$, there must be something I missed... – xavierm02 Jan 12 at 21:59 I still don't get it... "has no upper bound" $\equiv$ $\lnot$("has upper bound")$\equiv \lnot(\exists M, \forall N, A_N < M) \equiv \forall M \lnot(\forall N, A_N < M)\equiv \forall M ,\exists N, \lnot(A_N If$(a_n)$is bounded above then by the Monotone Convergence theorem,$(a_n)$converges which is a contradiction. Thus$(a_n)$is not bounded above by some$M>0$, that is $$\exists M>0\ \exists N\in \mathbb{N}:a_N>M$$ But as$(a_n)$is increasing, $$n\ge N\implies a_n>a_N>M$$ This holds for$M'<M$as well and so $$\forall R>0\exists N\in \mathbb{N}:\ n\ge N\implies a_n>R$$ and so$a_n\to +\infty\$.

Note that we only incorporate divergence to show un-boundedness. This is done by the Monotone Convergence Theorem for sequences

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