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

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SUMMARY

The discussion focuses on Proposition 2.3.15 from Houshang H. Sohrab's "Basic Real Analysis" (Second Edition), specifically the relationship between sequences \( t_n \) and \( s_n \). It is established that if \( t_n \leq s_n \), then \( \limsup_{n \to \infty} t_n \leq \limsup_{n \to \infty} s_n \), leading to the conclusion that \( \limsup_{n \to \infty} t_n \leq e \) since \( \lim_{n \to \infty} s_n = e \). The participants clarify that this follows from the definition of the limit superior (lim sup), reinforcing the importance of understanding this concept in real analysis.

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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 $$t_n \leq s_n$$ so that

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

My thoughts so far are as follows:

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

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

Peter
 

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

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

$$t_n \leqslant s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant e$$

(because $\displaystyle\limsup_{ n \to \infty } s_n = \lim_{ n \to \infty } s_n = e$).
 
Opalg said:
Slightly amend that to get

$$t_n \leqslant s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant e$$

(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 $$t_n \leqslant s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n$$ ...Peter
 
Peter said:
Have to check things ... certainly did not know (could not find a Proposition) that $$t_n \leqslant s_n$$

$$\Longrightarrow \limsup_{ n \to \infty } t_n \leqslant \limsup_{ n \to \infty } s_n$$ ...
It's true, though! See if you can prove it yourself, using the definition of limsup.
 
Opalg said:
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
 

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