MHB The Number e .... Another Issue/Problem Regarding Sohrab Proposition 2.3.15 ....

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The discussion revolves around understanding a specific proof in Houshang H. Sohrab's "Basic Real Analysis," particularly Proposition 2.3.15. The key question is how the inequality involving the sequence \( t_n \) leads to the conclusion that \( s_m = \sum_{ k = 0 }^m \frac{1}{ k! } \leq \text{ lim inf } (t_n) \). It is clarified that taking the limit inferior on both sides of the inequality \( t_n \geq 1 + 1 + \ldots \) with respect to \( n \) results in \( s_m \) being the limit of the right side, thus establishing the inequality. The discussion emphasizes the importance of understanding the behavior of sequences as \( n \) approaches infinity in the context of real analysis. Overall, the interaction aims to clarify the proof's implications and the reasoning behind the inequality.
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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 another issue/problem with the proof of Proposition 2.3.15 ...Proposition 2.3.15 and its proof read as follows:
View attachment 9076
In the above proof by Sohrab we read the following:

" ... ...Next, for any fixed $$m \in \mathbb{N}$$ and $$n \geq m$$, we have

$$t_n \geq 1 + 1 + \frac{1}{ 2! } \left( 1 - \frac{1}{ n } \right) + \ ... \ + \frac{1}{ m! } \left( 1 - \frac{1}{ n } \right) \ ... \ \left( 1 - \frac{m - 1}{ n } \right) $$so that letting $$n \to \infty$$ we get

$$s_m = \sum_{ k = 0 }^m \frac{1}{ k! } \leq \text{ lim inf } (t_n)$$ ... ... "
My question is as follows:

How does the statement:" ... ... for any fixed $$m \in \mathbb{N}$$ and $$n \geq m$$, we have

$$t_n \geq 1 + 1 + \frac{1}{ 2! } \left( 1 - \frac{1}{ n } \right) + \ ... \ + \frac{1}{ m! } \left( 1 - \frac{1}{ n } \right) \ ... \ \left( 1 - \frac{m - 1}{ n } \right) $$ ... " and letting $$n \to \infty$$ ...
... ... lead to the statement that ... $$s_m = \sum_{ k = 0 }^m \frac{1}{ k! } \leq \text{ lim inf } (t_n)$$ ... ... In other words ... can someone explain in some detail how/why $$s_m = \sum_{ k = 0 }^m \frac{1}{ k! } \leq \text{ lim inf } (t_n)$$ ... ...is true?

Help will be appreciated ...

Peter
 

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Hi Peter,

Peter said:
can someone explain in some detail how/why $$s_m = \sum_{ k = 0 }^m \frac{1}{ k! } \leq \text{ lim inf } (t_n)$$ ... ...is true?

Take liminf on both sides of the inequality $$t_{n}\geq 1+1+\ldots$$ with respect to $n$. Since the limit of the right side with respect to $n$ exists and is equal to $s_{m}$, this is the value of the liminf of the right side.
 
GJA said:
Hi Peter,
Take liminf on both sides of the inequality $$t_{n}\geq 1+1+\ldots$$ with respect to $n$. Since the limit of the right side with respect to $n$ exists and is equal to $s_{m}$, this is the value of the liminf of the right side.
Thanks GJA ...

Working on what you have suggested ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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