@StefanH.is this M not the same as L. Sequences can tend to a limit without exactly being theat so x$_{k}\leq M$ or L in my case.

@MichaelHardy, I tried to correct my proof above. Is it getting there now, or am i still missing something?

Why does $x_n> M +\varepsilon$ hold for every $n$? To me, you first claim is too fast. I would start proving that it is not bounded. (I think you still should show this? Can be done by reductio ad absurdum). Then somehow you know that a subsequence tends to infinity (why?). Why can you conclude that the whole sequence converges to infinity (Maybe the last two steps can be done in one step)

If a sequence has no convergent subsequence,you shouldn't have to prove that its not bounded above.If it were bounded you could pick a subsequence which converged to that so called upper bound. And that statement with $x_{n}$>M+$\epsilon$ holds for all n$\geq\textbf{N}$

An unbounded sequence does not necessarily tend to infinity. Think of $a_n= n$ for odd $n$ and zero for even $n$. Of course this sequence has a convergent subsequence.

But we are already assuming that it has no convergent subsequence, so this example doesn't apply.

What is THE upper bound? Do I find a convergent subsequence against any upper bound? I would simply apply Bolzano Weierstrass which states that a bounded sequence has a convergent subsequence. Then I've found a contradiction

of course this example does not apply. But you have to prove that this does not apply. You only know that the sequence is unbounded. This means: For every $M$ there exists some $n$ such that $x_n> M$. You gain no further information!

hey

hey...Why can't you just say x$_{n}$>M+$\epsilon$ this statement should still hold, and reverting back to the limit definition |x$_{n}-M|>$\epsilon$

That's the problem: "just say". You have to prove it. What you can say is that the sequence is unbounded. And unbounded means only what I said before.

But the clue is: Any subsequence is also unbounded:

What you should try to prove for the next step is the following. For each $M$ there exists an $N$ such that $x_n>M$ for every $n>N$. I would try to show this via contradiction

so if I told you a sequence doesn't converge, I would need to prove that it isn't bounded above?

$(-1)^n$ does not converge but is bounded. So I think you should prove this or as it is in this case mention that a bound would contradict Bolzano Weierstrass.

Now suppose we have a bounded sequence in R; so there exists a monotone subsequence, necessarily bounded. It follows from the monotone convergence theorem that this subsequence must converge. this would contradict that there are no convergent subsequences?

sure: since you found a convergent subsequence. But for me the Bolzano Weierstrass theorem just states that a bounded sequence has a convergent subsequence. So for me there is no monotonicity in there

and after this statement we can say the sequence is unbounded...and proceed like I was doing?

As said before: From that point on you know that for each $M$ there is at least one $x_n>M$. However this does not yet imply that the sequence tends to infty. For this you need to proof the following:

For each $M$ there exists an $N$ such that $x_n>M$ for every $n>N$. I would try to show this via contradiction

That's what i mentioned before

ok thx a lot

np

Last comment -1^n is also not convincing enough because we assumed that this sequence only contains positive real numbers.