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

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}

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

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.

• what is wrong in my proof? Mar 10 at 1:53
• Nice approach, which is the definition of $s_j$ Mar 10 at 1:56
• @JuanT it is just the sequence which you have provided. I have edited it in order to keep the notation. Mar 10 at 1:57
• 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? Mar 10 at 2:02
• @JuanT That is the point of the argument. Mar 10 at 2:04