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If $\lbrace S_n\rbrace_{n\in \mathbb{N}}$ is bounded and $\lim Sup \, S_n=\lim Inf\, S_n=s$ then $\lbrace S_n\rbrace_{n\in \mathbb{N}}$ converges to $s$.

Proof

Let us denote by $S$ be the set of subsequential limits, since $\lim Sup \, S_n=\lim Inf\, S_n=s$ then by definition of infimum. $\forall l\in S$ we have that $l\geq s$ and if $s^{\prime}$ is other number such that $l\geq s^{\prime}$ then $s^{\prime} <s$

And by definition of supremum $\forall l\in S$ we have that $l \leq s$ and if $s^{\prime}$ is other number such that $l\leq s^{\prime}$ then $s^{\prime} >s$.

From here $\forall l\in S$ $l=s$ and hence every subsequence of $\lbrace S_n \rbrace $ have limit $s$ hence $\lbrace S_n \rbrace $ is convergent sequence and in fact their limit is $s$

What is wrong with this proof? Any comment was very useful, thanks in advice

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HINT

Here it is another way to approach it for the sake of curiosity.

Notice that $m_{j}\leq S_{j} \leq M_{j}$, where \begin{align*} \begin{cases} m_{j} = \inf\{S_{n}\in\mathbb{R}\mid n\geq j\}\\\\ M_{j} = \sup\{S_{n}\in\mathbb{R}\mid n\geq j\} \end{cases} \end{align*}

Then apply the squeeze theorem.

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  • $\begingroup$ what is wrong in my proof? $\endgroup$
    – Juan T
    Mar 10 at 1:53
  • $\begingroup$ Nice approach, which is the definition of $s_j$ $\endgroup$
    – Juan T
    Mar 10 at 1:56
  • $\begingroup$ @JuanT it is just the sequence which you have provided. I have edited it in order to keep the notation. $\endgroup$
    – user0102
    Mar 10 at 1:57
  • $\begingroup$ The argue with this proof is , notice that $m_j<S_j<M_j$ as you define and then $\lim m_j \leq \lim S_j \leq \lim M_j$ and it means by squeeze theorem that $\lim inf m_j=s \leq \lim S_j \leq \lim Sup M_j=s$ and hence $\lim S_j=s$ as requiered? $\endgroup$
    – Juan T
    Mar 10 at 2:02
  • $\begingroup$ @JuanT That is the point of the argument. $\endgroup$
    – user0102
    Mar 10 at 2:04

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