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

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

Help will be much appreciated ...

Peter

 

[castor28]

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.

 

[Math Amateur]

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