Every bounded sequence converges Q) Prove or disprove : Every bounded sequence converges. (Make sure
to fully justify your answer. That is provide a proof if true otherwise
provide a counterexample and justify why your counterexample satisfy
the desired criteria.
Solution
My counter example is $a_n = (-1)^n $, I know I could just leave it as this but I want to prove that this diverges but I've never worked with a proof like that so could someone check it?
Firstly: {-1,1,-1,...} Therefore the sequence is bounded above by 1 and bounded below by -1. Therefore this sequence is bounded. 
Proof that this sequence diverges:
Assuming the contradiction that $\{a_n\}$ converges.
WTS: $\exists L \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|(-1)^n - L| < \epsilon$
Let $\epsilon = 1$
n is odd: if $n > N$, then $$|L+1| < 1$$
$$ \Leftrightarrow -2 < L < 0$$
n is even: if $n > N$, then $$|L-1| < 1$$
$$\Leftrightarrow 0 < L < 2$$
Thus $L \in (-2,0)$ and $L \in (0,2)$. Therefore this is a contradiction and $(-1)^n$ diverges. Therefore not all bounded sequences converge.
If this is correct I dont get why this actually proves it diverges, mainly this part: "$L \in (-2,0)$ and $L \in (0,2)$."
Could someone explain that? I was looking at a similar proof when writing this.
EDIT: Not a duplicate because I'm asking my additional proof is correct.
 A: Another proof: assume that $(a_n)$ converges. Then $|a_{n+1}-a_n| \to 0$. But we have:  $|a_{n+1}-a_n| =2$ for all $n$, a contradiction.
A: Yet another:
For any $L$,
$$\max(|L-1|,|L+1|)\ge1$$ so that you can't ensure $|L-(-1)^n|<\epsilon$.
A: You do not need assume something that you know is wrong and then expose a contradiction. Just recall the definition and show its prerequisities can't be satisfied.
The convergence is equivalent to, and in fact is defined by, the existence of a limit:

for a sequence, if (actually: iff) there exists a number $L$ which satisfies

for arbitrarily small margin allowed, almost all terms of the sequence get at least that close to $L$,

then we call $L$ a limit of the sequence, and we call the sequence 'convergent'.

Thus it's sufficient to show no candidate satisfies the requirements for a limit:

*

*for any $L\ne 1$ there exsist infinitely many terms equal $1$, that is $|L-1|$ apart from $L$, thus not arbitrarily close to $L$;

*and for $L=1$ there are infinitely many terms equal $-1$, thus not arbitrarily close to $L$,

to conclude: no candidate $L\in\mathbb R$ satisfies the definition's requirements, hence there is no limit, and the sequence is not convergent.
With no contradiction.
The same said @Yves Daoust before, and got the answer accepted, so this may be considered as duplicating the answer – but I wanted to show that basic mathematics can be done and explained without much formalism.
