Necessary and sufficient condition for the existence of $\lim_{n\to\infty} a_n$ Let {$a_n$} be a sequence of real numbers. Then what should be a necessary and sufficient condition for the existence of $\lim_{n\to\infty} a_n$?
P.S.- The condition should be in terms of existence of other limits, for example, it exists if $\lim_{n\to\infty} a_{2n}$ and $\lim_{n\to\infty} a_{3n}$ both exist, say...
 A: Cauchy's criterion is necessary and sufficient
$$\forall \epsilon > 0 \exists N: \forall n, m > N : |a_n - a_m| < \epsilon$$
Also, all subsequences of $a_n$ tends to the same limit (this is also necessary and sufficient)
A: As already mentioned, the existence of the limit of $a_n$ is a thing, the convergence is another matter entirely. I think the most useful result is the following: the limit
$
\begin{align}
\lim_{n \to +\infty} a_n
\end{align}
$
exists if and only if
$
\begin{align}
\liminf_{n \to +\infty} a_n = \limsup_{n \to +\infty} a_n
\end{align}
$
Note that in general
$
\begin{align}
\liminf_{n \to +\infty} a_n \leq \limsup_{n \to +\infty} a_n
\end{align}
$
This result is particularly useful since (for real successions) $\liminf$ and $\limsup$ always exist, even if the limit does not. Moreover, when they are equal, the limit is also equal to that common value.
P.S. See also here for limit superior and limit inferior.
A: A necessary, but not sufficient condition is that $a_n - a_{n + 1}$ goes to zero as n goes to $\infty$. (Usually this conditon is used as a necessary condition to see if a series converges: $\sum^\infty_{n=0}{b_n}$ can only converge if $b_n \rightarrow 0$. By setting $b_n = a_{n + 1} - a_n$ we get the above condition for sequences).
So, to recap: Cauchy's criteron is equivalent to the sequence having a limit, as Ant wrote in his answer.
Boundedness (first mentioned in Hamid's comment) and the difference going to zero are two necessary, but not sufficient conditions (they are not sufficient even if you combine them).
An example of a bounded sequence that doesn't converge is $a_n = sin(\frac{\pi}{2} n)$.
An example of a sequence where the difference goes to zero, but doesn't converge, is $a_n = \sum_{k=0}^n{\frac{1}{k}}$. An example that is bounded and where the difference goes to zero, but that still does not converge, is $a_n = sin(\frac{\pi}{2} n)\sum_{k=0}^n{\frac{1}{k}}$.
A: The boundedness is a Necessary condition but not a sufficient one. Also if a(n) converges, then it is a Cauchy sequence. But the converse is not true in general.
(this is true only in complete metric spaces.). Hence boundedness and cauchy criterion both are Necessary conditions for the convergence. 
A: First a short comment, the question asks for the existence of the limit, NOT about convergence. The two are not the same thing.
If you are looking for it in terms of limit, here is an idea:
Prove that 
$$\lim_n a_n$$
exists if and only if all of the following three limit exist:
$$\lim_n a_{2n} \,;\, \lim_n a_{2n+1} \,;\, \lim_n a_{3n}$$
And you only need the existence of the last limit to ensure that the first two limits are the same, you can replace the existence of $\lim_n a_{3n}$ by asking that the first two limits exist and are equal.
To prove this
Use first the fact that $a_{2n}$ and $a_{3n}$ have a common subsequence to show that they have the same limit. Repeat the argument with $a_{2n+1}$ and $a_{3n}$. Deduce that $\lim_n a_{2n} = \lim_n a_{2n+1}$.
Then prove using an $\epsilon, N$ argument that if $\lim_n a_{2n} = \lim_n a_{2n+1}=l$ then $\lim_n a_n=l$. This is pretty straightforward  proof.
The other implication is trivial.
