2
$\begingroup$

I have to prove that:

If a positive sequence of numbers $a_n$ is growing and bounded above, then the sequence $a_n$ converges to some finite value as $n$ increases.

Here is my 'proof' so far:

Suppose that a positive sequence $a_n$ is growing such that $a_n < a_{n+1}$, and bounded above by some number A such that $a_n \leq A, \forall n \in \mathbb{N}$.

We know from the supremum axiom that the set $M = {a_1, a_2, a_3, ...}$ not only has the upper bound A but a smallest upper bound which we can denote $\sup M = L$

Let $\epsilon > 0$ be a fixed and unknown number. We want to prove that $a_n \rightarrow L$, and we do this by finding a number $N$ so that

$$n>N \implies |a_n - L| < \epsilon$$

Opening up the absolute value sign we get: $$L - \epsilon < a_n < L + \epsilon$$

Since $L = \sup M$ of the set $M$ which contains all elements of the sequence $a_n$ and since $\epsilon > 0$ it follows that there must exist some $N$ such that $L-\epsilon < a_N$, which means that for all $n>N$ we have $L - \epsilon < a_n$.


Two questions: does my proof look alright so far? And secondly, I am supposed to find a suitable choice for $N$ which makes the implication $n>N \Rightarrow |a_n - L| < \epsilon$ true. How do I do this when I don't know what the sequence looks like?

$\endgroup$
2
  • $\begingroup$ Please note: my mathematics skills are not very advanced. Please keep this in mind :) Pardon any 'beginner' mistakes. $\endgroup$ Oct 4 '17 at 11:43
  • 1
    $\begingroup$ You may want to use the LaTeX/MathJax \mathbb 'blackboard font' modifier to the N letter to obtain \mathbb N → $\mathbb N$ symbol for the set of natural numbers. $\endgroup$
    – CiaPan
    Oct 4 '17 at 12:06
3
$\begingroup$

There is an error of logic in your proof.

The statement $$n>N \implies |a_n - L| < \varepsilon$$ is exactly what you want to prove to show that $a_n$ converges to $L$. What you have done is use this statement to prove the very same statement.

There is no way you can find a fixed relationship between $N$ and $\varepsilon$ by the way, as you have a generic and unspecified monotone increasing sequence. You can however do the proof nonetheless.

Hint: There are many different ways of proving this, but one fundamental approach could be: for a particular $\varepsilon$ and some integer $m>1$, consider the set $[L-\varepsilon, L-\varepsilon/m)$ and $[L-\varepsilon/m, L]$. Where do an infinite number of elements $a_i$ reside? Then increase $m$.

$\endgroup$
4
  • $\begingroup$ Sorry, I'm confused now. Where does the small m come from? $\endgroup$ Oct 4 '17 at 12:15
  • $\begingroup$ Just edited to make it more clear. $m$ is a number that you increase and hence will decrease the little $interval $[L-\varepsilon/m, L]$ $\endgroup$ Oct 4 '17 at 12:33
  • $\begingroup$ I'm still confused, sorry. I really like the idea of the sets but I'm having trouble wrapping my head around it. In the sets $[L- \epsilon, L - \epsilon /m)$ and $[L - \epsilon /m, L]$ where m is small, the second set will contain an infinite number of elements(?). But as m gets larger and larger, the sets will approximate to [L - \epsilon /m, L)$ and $[L, L]$. So the first set will contain infinite terms? $\endgroup$ Oct 4 '17 at 19:32
  • 1
    $\begingroup$ So we don't want to consider small $m$ at all, we only want to consider larger and larger $m$. The second statement you make about $m$ getting large is not quite right. The point is that as $m$ gets large, $\varepsilon/m$ becomes very very small, so $L-\varepsilon/m$ is indeed getting closer to $L$ (but doesn't touch yet). Now, what you want to say, using the logic of the monotone increasing sequence that is approaching $L$, is that an infinite number of $a_i$ reside in this tiny interval, no matter how large $m$ gets... $\endgroup$ Oct 5 '17 at 8:51
1
$\begingroup$

The proof looks basically ok. The idea is certainly correct.

I would write, instead of "let $\epsilon$ be a fixed and unknown number", something like "choose an arbitrary $\epsilon>0$". I would also perhaps point out that by definition of supremum we automatically and trivially have $a_n <L+\epsilon$ for all $n$. And I'd similarly point out that the last conclusion (if it works for $a_N$, it works for every $a_n$ with $n>N$) follows from the fact that the sequence is non-decreasing. The last two observations are probably superfluous if writing "professionally", but it's probably better to write them down for a homework.

To answer your second question: how can I know which N is "sufficiently large" for a given $\epsilon$? Well, as we said, if $a_N > L-\epsilon$, you are ok for every $n>N$ too, since the sequence is non-decreasing. Special sequences might sport special properties that allow you to determine a "sufficiently large N" but, in general, determining $N$ might require you to try $a_1$, $a_2$ etc. until you find something exceeding $L-\epsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.