1
$\begingroup$

Im trying to prove that if $\lim \limits_{n \to \infty} a_n=\infty$, then $a_n$ has a minimum.

All I could think about is the definition of convergence to infinity, which says that:

for each $M>0$ there is such $N$ , s.t. for each $n>N$ $a_n>M.$

and I'm trying to use this definition by mixing it with the intuition that a convergent series to "$\infty$", can't just jump back to "$-\infty$" after passing certain $N$, then at some point, it must have a startting point that its $NOT$ "$-\infty$" from which it increased to "$\infty$". and that starting point must be the minimum.

how do I turn these words into a mathematical form? any kind of hints would be appreciated.

$\endgroup$
0

5 Answers 5

3
$\begingroup$

Your reasoning about "$-\infty$" is not very meaningful as written, though it may encode a valid intuituion. In order to make it precise, you need to think about precise boundaries, not just handwavy "$-\infty$".

What I would do:

In the definition of convergence, set $M=a_0$. This tells you that there are at most finitely many terms in the series that are less than or equal to $a_0$ (and there is at least one such, namely $a_0$); one of these must be the minimum, and thus a minimum for the whole sequence.

$\endgroup$
2
  • $\begingroup$ yeah, it makes alot of sense. but I needed something to strengthen the fact that the terms who are less than $a_0$ are finite. I guess it is guaranteed by the definition of convergence to infinity. $\endgroup$
    – F1sargyan
    Commented Nov 7, 2015 at 17:32
  • $\begingroup$ @F1sargyan: But that's exactly what the definition of convergence gives you: The only $a_n$s that can possibly be $\le M$ are those with $n\le N$, so there can be at most $N+1$ of them, which is finitely many. $\endgroup$ Commented Nov 7, 2015 at 17:43
1
$\begingroup$

Suppose a sequence $(a_i)_{i\in\mathbb{N}}$ does not have a minimum. That means, for each term $a_i$, there is some term $a_j$ such that $a_j<a_i$.

EXERCISE: show that we can assume $j>i$ - that is, for each term $a_i$ there is a later term $a_j$ which is less than $a_i$.

So now we get a subsequence of our original sequence: start at $b_0=a_0$ . . .

  • Find some later term $b_1$ which is $<b_0$;

  • Find some later term $b_2$ which is $<b_1$;

  • Etc.

Then we have $b_0>b_1>b_2> . . . $ Why does this contradict the assumption that the $a_i$s go to $\infty$? (HINT: Think of some $M$ which the sequence $(a_i)_{i\in\mathbb{N}}$ never stays bigger than . . .)

$\endgroup$
1
$\begingroup$

Since $a_n\to \infty $, there is an $N\in\mathbb N$ s.t. $a_n>a_0$ for all $n>N$. In particular, $$a_n\geq \min\{a_0,...,a_N\}$$ for all $n$.

$\endgroup$
1
$\begingroup$

You are on the right track, a direct proof is as follows, for $a_1$, there exists $N$ such that for all $n > N$, $$a_n > a_1. \tag{1}$$ Now let $a_{i_0}$ be the minimum of the set $\{a_1, \ldots, a_N\}$, which exists since the set is finite. Then $a_{i_0}$ is the minimum of the whole sequence $\{a_n\}$ since by $(1)$ for all $n > N$, $$a_n > a_1 \geq \min(a_1, \ldots, a_N) = a_{i_0}.$$

$\endgroup$
1
$\begingroup$

Sketch of Proof: There is an $N$ such that $a_n>a_1$ whenever $n>N$.

Among the first $N$ elements of the sequence, there is a minimum, and this must be the overall minimum of the sequence.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .