Is every bounded sequence convergent? It's true, that every convergent sequence is bounded, but is every bounded sequence convergent?
 A: No. Take the sequence $a_n=i^n$ in the complex plane. It is bounded since it is contained in the circle $|z|\leq \sqrt{2}$. However it doesnt converge since its terms alternate depending on $n$ modulo 4.
A: No, there are many bounded sequences which are not convergent, for example take an enumeration of $\mathbb Q\cap(0,1)$.
But every bounded sequence contains a convergent subsequence.
A: No. For example, take the sequence
$$a_n=\begin{cases}
0 & \text{ if }n\text{ is even}\\
1 & \text{ if }n\text{ is odd}
\end{cases}$$
It is bounded because it stays inside the interval $[0,1]$, but it has no limit.
Intuitively, you shouldn't expect that bounded $\implies$ convergent, because even if the terms of a sequence stay in some general area, doesn't mean that all of its terms must always be getting closer and closer to each other (which is what the notion of Cauchy sequence captures; a sequence in $\mathbb{R}$ or $\mathbb{C}$ is convergent $\iff$ it is Cauchy).
However, as user amWhy points out in their answer, every bounded sequence contains a convergent subsequence; in other words, we can pick out some terms of the sequence that are getting closer and closer to each other (even if they aren't getting closer to all the terms in the original sequence).
A: Hint: $\quad$Consider the sequence $\{a_n\},\;a_n = (-1)^n\,$
It is bounded in $[-1, 1]\; ($ indeed, $a_n \in \{-1, 1\}\; \forall a_n\in \{a_n\}),\;$ but $\;\lim_{n\to \infty} (-1)^n\;$ does not exist.
Note: it is true that every bounded sequence contains a convergent subsequence, and furthermore, every monotonic sequence converges if and only if it is bounded. 

Added
See the entry on the Monotone Convergence Theorem for more information on the guaranteed convergence of bounded monotone sequences.
A: Look at the sequence : -1,1,-1,1,...
A: Let $\{X_n\}$ be a convergent sequence, converging to $L$, then corresponding to $\epsilon=1$, there exist $N_0\in \Bbb N$ such that $|X_n-L|<1 \forall N\geqslant N_0$. 
