# Doubt regarding: Does existence of limsup of a sequence $x_n$ imply its convergence? Here is the proof that it does:

### Let's consider only bounded sequences.

Let ($$x_n$$) be a bounded sequence then I am using the following definition for limsup of the sequence ($$x_n$$):
limsup($$x_n$$)= inf $$V$$=inf{$$v: x_n >v$$ for at most a finite natural number $$n$$}=$$x^*$$, say. Consider the following: $$(a) \Rightarrow (b)$$: Since $$x^*$$ is inf $$V$$ hence for any $$\epsilon \gt 0 , \exists v\in V:$$ $$x^* \le v\lt x^*+\epsilon \implies x^*\in V \implies$$ there can be atmost a finite number of $$n\in N$$ for which $$x^*+\epsilon\lt x_n$$. $$\forall \epsilon \gt 0$$ $$x^*-\epsilon \notin V$$. Hence $$x_n \gt x^* -\epsilon$$ for infinitely many $$n$$.

Now my claim is that $$(b)$$ implies that $$x_n$$ is convergent, below is the proof:
From $$(b)$$ above, let $$n=K$$ (this K either exists or doesn't exist because of "atmost" condition) be the number for which $$x^*+\epsilon\lt x_n$$.
(A) $$\forall n\ge K$$, we have $$x_n \lt x^* +\epsilon$$
(B) Now we know that $$x_n \gt x^* -\epsilon$$ for infinitely many $$n \implies \exists M \in N : \forall n \ge M$$, we have $$x_n \gt x^* -\epsilon$$
From (A) and (B), for $$n \ge L$$=sup{$$K,M$$}, we have
$$x^* -\epsilon \lt x_n \le x^* +\epsilon \Rightarrow |x_n-x^*|\lt \epsilon \implies$$ $$lim (x_n)=x^*$$

I know that if $$X_n=$$ sup{$$x_k: k\ge n$$}, then $$lim (X_n)=x^*=$$ inf{$$X_m, m\in N$$}. That's why the proof above doesn't seem right to me. I am struggling to know what is wrong in the above proof. Please help. Thanks in advance.

• $x_n=\sin{n}$ has a lim sup of 1 but it does not converge. – Paul Mar 27 at 18:42
• @Paul, even for $x_n =1+(-1)^n$ limsup exists but $x_n$ doesn't converge. But the problem is what went wrong in my proof then? – Koro Mar 27 at 18:57
• see my answer below. – Paul Mar 27 at 18:58

The key is that the 2 statements $$x^*+\epsilon and $$x^*-\epsilon do not cover all the possibilities for the $$x_n$$. In particular, you don't know anything about the case where $$x^*-\epsilon >x_n$$. This could be finite or infinite. You can look at the example of $$x_n=\sin{n}$$ for an example where both regions contain an infinite number of $$x_n$$.

• +1. From what I understand you mean that (P): $x_n \gt x^* -\epsilon$ for infinitely many $n$ means (P) may be true for some subsequence of (n) but for some other subsequnce of (n), $x_n \lt x^* -\epsilon$ and therefore my statement (B) is wrong. – Koro Mar 27 at 19:07
• @Koro, that is correct. – Paul Mar 27 at 19:09
• you might want to change $x_n=sinx$ to $x_n=sin n$ in your answer. – Koro Mar 27 at 19:11
• Ah, yes, missed that! – Paul Mar 27 at 19:13

The implication you made in statement (B) is false. Just because there are infinitely many terms of the sequence that satisfy that property does not mean that they must all be consecutive.

• Thanks a lot. I understood. – Koro Mar 27 at 19:07

A statement being true for infinitely many terms doesn't mean that it is true for all but finitely many terms. Any bounded sequence has a $$\limsup$$, but most of them don't converge.

• Yeah right. Got it thanks +1. – Koro Mar 27 at 19:08