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I am studying for my first exam in Analysis 1 and one I have to know how to prove that a monotonic sequence in $\mathbb{R}$ is convergent if and only if it is bounded.

I start by proving the first implication, thus that monotonic and convergent => bounded. This is easy as I am allowed to use an earlier proof from my book to say that any convergent sequence in $\mathbb{R^n}$ is bounded.

However the other implication gets a little bit trickier. I know that I have to prove that monotonic and bounded => convergent.

I first start by assuming that the sequence $(a_k)_{k=1}^\infty$ is increasing. Thus the set $\{a_k : k \in \mathbb{N}$} of the values of the sequence is non-empty and bounded and therefore it has a well-defined supremum.

I let $a=sup\{a_k : k \in \mathbb{N}\}$ and claim that the sequence converges to a. Thus I let $\epsilon > 0 $ and use the definiton of convergence $\forall \epsilon > 0 \exists N \in \mathbb{N} : k \geq N \rightarrow ||a_k-a||< \epsilon$ which I negate for a contradiction, ie. $\exists \epsilon > 0 \forall N \in \mathbb{N} : \exists k \geq N$ and $a-a_k \geq \epsilon $.'

Then I create a new, strictly positive sequence $(k_j)_{j=1}^\infty$ with $a-a_{kj} \geq \epsilon $ and $j \in \mathbb{N}$ but why does it need to be strictly positive and not just positive. Does this make a difference?

I now let $N=1$ to obtain that $k_1 \geq 1$. By constructing $k_{j-1}$ it immediately follows that $k_j \geq k_{j-1} + 1 > k_{j-1}$. Is this because that I have just seen that for $N=1$ that $k_1 \geq 1$ which means that $k_j \geq k_{j-1} +1$ as $j \in \mathbb{N}$? I am not sure how to explain it ..

As I now have constructed a strictly positive sequence I know that $\forall k \in \mathbb{N}$ I can find a $j \in \mathbb{N}$ such that $k \leq k_j$ but why is this? My friend tried to explain it is because that I first pick a $k$ and then afterwards I can pick a $j$ and therefore $k \leq k_j$. Likewise this means that, by induction, $a_k \leq a_{kj}$. Can this also be explained by saying that the sequences are positive? Or only by induction? I dont have time for that if I have to present this subject for my professor.

Now I know this means that $a-a_k \geq a-a_{kj} \geq \epsilon $ but then $a-a_k \geq \epsilon $ and solving for $a_k$ we obtain that $a-\epsilon \geq a_k$ implying that this is an upper bound for the sequence contradicting the fact that $a = \sup \{a_k \ k \in \mathbb{N}$} which completes the proof.

I hope you can clarify my questions.

Thanks in advance.

Arturo Magidin
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Mathias
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1 Answers1

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Do it directly: the fact that $\sup\{a_n:n\in\mathbb N\}=a$ implies two things:

$1).\ $ that $a_n\le a$ for all integers $n$

$2).\ $ there is an integer $N$ such that $|a-a_N|<\epsilon$.

But $(a_n)$ is increasing , so if $n>N$ then $a_N\le a_n\le a$, so in fact $|a-a_n|=a-a_n<\epsilon$, which means that $\underset{n\to \infty}\lim a_n=a.$

Matematleta
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