P-series .... Sohrab, Proposition 2.3.12 .... ....

[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 2.3.12 ...

Proposition 2.3.12 and its proof read as follows:

View attachment 9051
In the above proof by Sohrab we read the following:

" ... ... Now if \(\displaystyle p \leq 1\), then \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ... "My question is ... how do we know this is true ... ?

Can someone please demonstrate how to prove that if \(\displaystyle p \leq 1\), then \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ...
Help will be much appreciated ...

Peter

 

[Olinguito]

" ... ... Now if \(\displaystyle p \leq 1\), then \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ... "My question is ... how do we know this is true ... ?

Can someone please demonstrate how to prove that if \(\displaystyle p \leq 1\), then \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ...

If $p\le1$ then $q=p-1<0$ and so $n^q\le1$.

 

[Math Amateur]

If $p\le1$ then $q=p-1<0$ and so $n^q\le1$.

Thanks for the reply Olinguito ...... but ... I do not follow ... can you give some more details please ...Peter

 

[Olinguito]

If $q$ is negative, then $n^q=e^{q\ln n}\le1$ for any positive integer $n$, is that not so?

 

[Math Amateur]

If $q$ is negative, then $n^q=e^{q\ln n}\le1$ for any positive integer $n$, is that not so?

Hi Olinguito ...

It seems plausible that if \(\displaystyle q \lt 0\) then \(\displaystyle n^q \lt 1\) ... but how do we (formally and rigorously) prove it ... especially worrying, intuitively speaking, are those values \(\displaystyle -1 \lt q \lt 0\) ...

... and then how do we demonstrate(formally and rigorously) that \(\displaystyle n^q \lt 1\) implies \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ..*** EDIT *** Can now see that \(\displaystyle n^q \lt 1\) implies \(\displaystyle 1 / n^p \geq 1 / n \ \forall \ n \in \mathbb{N}\) ... indeed quite straightforward ... getting late here in Tasmania ...:(...
Sorry if I'm being a bit slow ...

Peter