Relationship between limsup, liminf and limit. If you want to show that a sequence $(a_{n})$ in $\mathbb{R}$ is convergent, when is it sufficient to show that there is a number $b\in\mathbb{R}$ such that
$$ \liminf a_{n} \geq b \geq \limsup a_{n}$$
In particular, I have a situation where my sequence is bounded and I wanted to use this approach, but I'm not sure I really understand what is going on and why or if this works.
Thanks for any illumination!
 A: $$\forall n\in\mathbb{Z}^+[\inf\{a_m|m\geq n\}\leq\sup\{a_m|m\geq n\}]$$
Thus:
$$\lim\inf a_n\leq\lim \sup a_n$$
Now if you can show that $\lim\sup a_n\leq\lim\inf a_n$, then this inequality and the previous one show that $\lim\inf a_n=\lim\sup a_n$
The last equation can be used to show that $\lim a_n$ exists
A: Geometrically, $\lim \inf$ is the leftmost (or smallest) accumulation point of a sequence, while $\lim \sup$ is the rightmost (or biggest) one. If you can prove that the one on the left is in fact bigger than or equal to the one on the right, then they coincide. Thus, your sequence is convergent, by Weierstrass theorem, since it only has one accumulation point. (Some take it as a definition.)
A: One way to think about this is that the $\liminf a_n$ is the smallest subsequential limit (the smallest limit of any subsequence of $a_n$), and the $\limsup a_n$ is the largest subsequential limit. So if you can show that $\liminf a_n \geq b \limsup a_n$, then you will have in fact shown $\liminf a_n = b = \limsup a_n$ because we will always have $\liminf a_n \leq \limsup a_n$. (The Wikipedia article is not bad reading.)
But sometimes it is not clear what the actual limit of the sequence will be, so it is nice to have other criteria for convergence that do not require explicitly finding the limit of the sequence. Have you encountered the Cauchy criterion yet?
In fact, a sequence $a_n$ converges iff $\liminf a_n = \limsup a_n$ iff $a_n$ satisfies the Cauchy criterion (and, if $a_n$ converges, then $\liminf a_n = \limsup a_n = \lim a_n$).

I have a situation where my sequence is bounded

Another possible option: If the sequence is bounded and also monotonic, then it will converge. But your sequence may not have this condition.
A: It is always sufficient to know such a $b$ exists. Since $\liminf a_n \le \limsup a_n$ always holds, if a $b$ as in the question exists then in particular holds that $\limsup a_n \le \liminf a_n$. But then the equality $\limsup a_n = \liminf a_n $ holds. Now, a sequence converges if, and only if, its limsup is equal to its liminf. So, the sequence converges. 
