# 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...

• since every convergent sequence is bounded, then boundedness is a Necessary condition for the convergence. However, a(n) converges iff it is a Cauchy sequence.(in this case, of course) – Fermat Apr 26 '14 at 19:17
• The condition I wrote (the second one) applies precisely to $a_{2n}, a_{3n}$ and in general to $a_{f(n)}$ – Ant Apr 26 '14 at 20:29

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)

• $\lim_n n$ exists but $n$ is not Cauchy ;) – N. S. Apr 26 '14 at 19:37
• I interpreted that as "converges", basically the limit is finite.. Otherwise you're right, though the second condition is still valid.. Thanks for pointing that out :-) – Ant Apr 26 '14 at 19:50

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.

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 $a_n = sin(\frac{\pi}{2} n)\sum_{k=0}^n{\frac{1}{k}}$.

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.

• The boundedness is not necessary for the existence of the limit. It is only necessary for convergence, but what if the limit exists and is infinite? – N. S. Apr 26 '14 at 19:42

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.

• "you only need the existence of the last limit to ensure that the first two limits are the same" provided that the first two limits exist, otherwise it's not true. Take the sequence: $a_0,0,1,1,0,1,1,0,1,1,\dots$. In essence, $a_{3n}=1$ ($a_0=1$). Now, $\lim_{n} a_{3n}=1$, but $\lim_n a_{2n}$ doesn't exist. Having a common subsequence is not enough to conclude that they have the same limit; the limit, first and foremost, must exist. You have to slightly change your proof in order to exclude degenerate cases. Or simply use $\liminf$ and $\limsup$ (which I prefer). – PseudoRandom Apr 27 '14 at 10:50
• @PseudoRandom You only read half of the paragraph, the proposition before states exactly what you said. And the proposition you quoted sais something different than what you are implying: it sais that for the proof this condition is ONLY needed to guarantee that the three limits are the same, it is not needed for anything else. – N. S. Dec 21 '14 at 14:26
• @PseudoRandom Let me quote what my post sais: "Prove that $$\lim_n a_n$$ exists if and only if all of the following three limit exist: $$\lim_n a_{2n} \,;\, \lim_n a_{2n+1} \,;\, \lim_n a_{3n}$$ " So I don't understand your complain. The inverse is trivial, and for the converse, the existence of the three limits is part of the statement... – N. S. Dec 21 '14 at 14:27
• There are two main complains: the first is the wording of the sentence I quoted (you should explicitly add "provided that the first two limits exist"). The second, and more important, is about the hypothesis you choose. In my humble opinion, it seems conceptually more clear, more elegant and more straightforward to ask that the first two limits exist and are equal. Without resorting to the existence of limit of $a_{3n}$, just like you suggested (but didn't do). The natural consideration is: if the first two limits exist, cheking if they are equal seems easier and cleaner. – PseudoRandom Jan 2 '15 at 22:25
• My vote is now locked, but if you do a simple edit, it will unlock and I can change it. – PseudoRandom Jan 2 '15 at 22:27