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Okay so here I've tried proving that :

Let $x_n$ be a sequence, if $\limsup{x_n}=\liminf{x_n}$ then $x_n$ converges.

Proof. By definition, $\limsup{x_n}$ and $\liminf{x_n}$ are limits of subsequences that are formed from $x_n$. Let these subsequences be $a_n$ and $b_n$ such that:

$$\lim{a_n}=\limsup{x_n}=a \\ \lim{b_n}=\liminf{x_n}=b$$

Okay let's analyse this :

There exists and $ɛ_1$ such that $|a_n-a|<ɛ_1$ for $n>N^{'}$

There exists and $ɛ_2$ such that $|b_n-b|<ɛ_2$ for $n>N{''}$

Let

$$N=max(N{'},N{''})$$

so $|a_n-a|+|b_n-b|<ɛ_1+ɛ_2$ for $n>N$

By triangle inequality ::

$$|a_n+b_n-(a+b)|<ɛ_1+ɛ_2$$

So we have found that when we add this subsequences together, they converge to $a+b$, what if $a=b$ Then : $$|a_n+b_n-(2a)|<ɛ_1+ɛ_2$$

So both $a_n$ and $b_n$ have infinite traces around the point $a$ Since their trace is a subset of the actual sequence $x_n$ we can considered that they are approaching the open ball $B(a,ɛ)$. when they approach to the ball, for $n>N$ sequences equal out $a_n=b_n$ and we can also say that

$$a_n=b_n=x_n$$ for $$n>N$$

So that when $\liminf=\limsup$ the sequence $x_n$ converges.

What do you think about my proof? IS it correct? What else should I add, what should I adopt? I'm new to analysis so help me out please.

Thanks.

1 Answers1

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$l_n=\inf_{k \ge n} a_k \le a_n \le \sup_{k \ge n} a_k=u_n$. Note that $l_n$ is non decreasing and $u_n$ is non increasing.

Let $l = \liminf_k a_k $ and $ u = \limsup_k a_k $.

Suppose $l=u$ and $\epsilon>0$. Choose $N$ such that $u_n-l_n <\epsilon$ for $n \ge N$. Then $0 \le a_n-l \le a_n - l_n \le u_n - l_n < \epsilon$ for all $n \ge N$. In particular, this shows that $a_n \to l$.

copper.hat
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