# Prove that if $a_n$ contains no convergent subsequences, then |$a_n$| approaches infinity, as n approaches infinity.

I first approached the problem using contradiction.

Suppose that |$a_n$| does not approach infinity. Case 1: The sequence will converge to some L, and thus be bounded. Case 2: The sequence will neither converge nor diverge, thus it will be bounded. (I'm having trouble explaining case 2)

For both cases, we can use the Bolzano-Weiestrass Theorem, thus |$a_n$| has a convergent subsequence. (How can I show that $a_n$ will also have a convergent subsequence?)

This is a contradiction to the given statement, thus |$a_n$| will approach infinity.

This was how I approached the problem, but there's a few problems with it. Please let me know if there is a better way or how I can improve mine.

OR

Again using contradiction, I can say that there exists an M > 0 for all N in the natural numbers, where there exists n > N such that we have |$a_n$| ≤ M.

Let n = $a_ni$. Then |$a_ni$| ≤ M. Thus -M < $a_ni$ < M. This sequence is bounded, thus, by the Bolzano-Weiestrass theorem, it has convergent subsequences. CONTRADICTION

• Your second approach is the better one imo. You just need to polish it up a little bit.
– DRF
Nov 20 '15 at 22:56

Suppose by contradiction that $|a_n|$ doesn't diverge to $\infty$. This means that there exists $M>0$ such that, for all $n$, there exists $m>n$ with $|a_m|\le M$.

So, start with defining $n_0$ so that $|a_{n_0}|\le M$. Then take $n_1>n_0$ such that $|a_{n_1}|\le M$ and so on. More precisely, if $n_k$ has been defined, take $n_{k+1}>n_k$ so that $|a_{n_{k+1}}|\le M$.

The subsequence $(a_{n_k})$ has its values in $[-M,M]$, so it is bounded and therefore it has a convergent subsequence, which is also a convergent subsequence of the original sequence.

Hint: what $|a_n|$ tends to $\infty$ as $n$ tends to $\infty$ means is that for any $x > 0$, there is an $N$ such that for any $n > N$, $|a_n| > x$. The negation of this condition is that there exists an $x > 0$, such that for any $N$, there is an $n > N$, such that $|a_n| \le x$. So if $|a_n|$ does not tend to $\infty$ as $n$ tends to $\infty$, take the $x > 0$ given by the negation of the condition and take $n = 1$ to get $n_1 > 1$ such that $|a_{n_1}| \le x$; now take $N = n_1$ to get an $n_2 > n_1$ such that $|a_{n_2}| \le x$. Continuing in this way you get a bounded subsequence of $a_n$, which will have a convergent subsequence by Bolzano-Weierstrass.

• Why was this down voted? What on earth is wrong with it? Nov 23 '15 at 20:13

One-point compactify $\mathbb{R}$. It is clear that $|x_n| \rightarrow \infty$ (in the usual "calculus sense") iff $x_n \rightarrow \infty$ in the one-point compactification

Given any subsequence of the sequence, it must have a convergent sub(sub)sequence, since it is on a compact set. Hence, since it does not converge to anyone in $\mathbb{R}$, this sub(sub)sequence converges to $\infty$. We proved that every subsequence has a sub(sub)sequence converging to a fixed point (namely, $\infty$). Hence, the sequence converges to this point ($\infty$).

Hint.: If $|a_n|$ doesn't approach infinity (i.e. it is not true that $\forall M>0\; \exists N\; \forall n \;[(n>N)\implies (|a_n|>M)]$) then there must exist a bounded subsequence. To prove that you just need to negate the statement in parentheses.

Once you have a bounded subsequence you can use some theorem I'm sure. If you don't know the theorem you might have your work cut out for you since you need completeness of the reals to get the result.

• The sequence $0,1,0,1,\ldots$ diverges. The term diverge implies nothing about the sequence other than the fact that it does not converge. Nov 21 '15 at 6:49
• @MattSamuel Hmm fair enough my bad. Should have checked my definitions. Should be fixed now.:) Ty for pointing it out.
– DRF
Nov 21 '15 at 7:51