The Number e .... Another Question Regarding Sohrab Proposition 2.3.15 ....

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with an another aspect of the proof of Proposition 2.3.15 ...

Proposition 2.3.15 and its proof read as follows:
View attachment 9072
In the above proof by Sohrab, we read the following:

" ... ... It follows that \(\displaystyle t_n \leq s_n\) so that

\(\displaystyle \text{ lim sup } ( t_n ) \leq e\) ... ... "
Can someone please explain exactly how/why \(\displaystyle t_n \leq s_n \Longrightarrow \text{ lim sup } ( t_n ) \leq e\) ... ... ?

My thoughts so far are as follows:

\(\displaystyle t_n \leq s_n\)\(\displaystyle \Longrightarrow \lim_{ n \to \infty } t_n \leq \lim_{ n \to \infty } s_n\) \(\displaystyle \Longrightarrow \lim_{ n \to \infty } t_n \leq e\)But ... how/why can we conclude that \(\displaystyle \text{ lim sup } ( t_n ) \leq e\) ... ... ?

***EDIT*** In the above thoughts I have wrongly assumed that we know, without further analysis, that \(\displaystyle (t_n)\) is convergent ... Help will be appreciated ... ...

Peter

 

[Opalg]

Can someone please explain exactly how/why \(\displaystyle t_n \leq s_n \Longrightarrow \text{ lim sup } ( t_n ) \leq e\) ... ... ?My thoughts so far are as follows:

\(\displaystyle t_n \leq s_n\)\(\displaystyle \Longrightarrow \lim_{ n \to \infty } t_n \leq \lim_{ n \to \infty } s_n\) \(\displaystyle \Longrightarrow \lim_{ n \to \infty } t_n \leq e\)But ... how/why can we conclude that \(\displaystyle \text{ lim sup } ( t_n ) \leq e\) ... ... ?

Slightly amend that to get

\(\displaystyle t_n \leqslant s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant e\)

(because $\displaystyle\limsup_{ n \to \infty } s_n = \lim_{ n \to \infty } s_n = e$).

 

[Math Amateur]

Slightly amend that to get

\(\displaystyle t_n \leqslant s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant e\)

(because $\displaystyle\limsup_{ n \to \infty } s_n = \lim_{ n \to \infty } s_n = e$).

Thanks Opalg ...

Reflecting on what you have written ...

Have to check things ... certainly did not know (could not find a Proposition) that \(\displaystyle t_n \leqslant s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n\) ...Peter

 

[Opalg]

Have to check things ... certainly did not know (could not find a Proposition) that \(\displaystyle t_n \leqslant s_n\)

\(\displaystyle \Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n\) ...

It's true, though! See if you can prove it yourself, using the definition of limsup.

 

[Math Amateur]

It's true, though! See if you can prove it yourself, using the definition of limsup.

Thanks for the help, Opalg ...

Yes, checked that out ... indeed basically follows from definition of lim sup ...

Thanks again

Peter