The Number e .... Sohrab Proposition 2.3.15 ....

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The discussion centers on Proposition 2.3.15 from Houshang H. Sohrab's "Basic Real Analysis" (Second Edition), specifically the proof that $$1 / n! \leq 1 / (2 \cdot 3^{n - 2})$$ for all $$n \geq 3$$. The proof is established by recognizing that the factorial $$n!$$ is the product of $$n$$ factors, where each factor after 2 is at least 3. This leads to the conclusion that $$n! \geq 2 \cdot 3^{n - 2}$$, allowing the inequality to be validated by taking the inverse.

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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 aspect of the proof of Proposition 2.3.15 ...

Proposition 2.3.15 and its proof read as follows:
View attachment 9052
In the above proof by Sohrab we read the following:

" ... ...Since $$1 / n! \leq 1 / (2 \cdot 3^{ n - 2 } )$$ for all $$n \geq 3$$ ... ... "My question is ... how do we know this is true ... ?

Can someone please demonstrate how to prove that $$1 / n! \leq 1 / (2 \cdot 3^{ n - 2 } )$$ for all $$n \geq 3$$ ... ...

Help will be much appreciated ...

Peter
 

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Hi Peter,

$n!=1\cdot2\cdot3\cdots n$ is the product of $n$ factors. As each of the $n-2$ factors after $2$ is at least equal to $3$, we have:
$$
n!\ge1\cdot2\cdot3^{n-2}=2\cdot3^{n-2}
$$
and the result follows by taking the inverse.
 
castor28 said:
Hi Peter,

$n!=1\cdot2\cdot3\cdots n$ is the product of $n$ factors. As each of the $n-2$ factors after $2$ is at least equal to $3$, we have:
$$
n!\ge1\cdot2\cdot3^{n-2}=2\cdot3^{n-2}
$$
and the result follows by taking the inverse.

Hmmm ... easy when you see how ... :) ...

Thanks for the help castor28 ...

Peter
 

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