Convergence of Geometric Series .... Sohrab, Proposition 2.3.8 .... ....

[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 2.3.8 ...

Proposition 2.3.8 and its proof read as follows:
View attachment 9048
In the above proof by Sohrab we read the following:

" ... ... Now, if $$\mid r \mid \lt 1$$ then (by Exercise 2.2.7) $$\lim (r^{ n + 1}) = 0$$ and $$( \ast )$$ follows at once ... ... "
Now it appears that Sohrab is referring to Part 2 of Exercise 2.2.7 (see below) that states:

" ... ... if $$0 \lt b \lt 1$$, then $$\lim ( b^n ) = 0$$ ... ... "Now my problem is we are not given $$0 \lt r \lt 1$$ ... but instead we are given $$\mid r \mid \lt 1$$ ... .. so my question is ... what do we do here ...?
Perhaps we put $$b = \mid r \mid$$ ... but this allows us to conclude $$\lim ( \mid r \mid^n ) = 0$$ ...... can we conclude from here that $$\lim (r^{ n + 1}) = 0$$ ...... by arguing that $$\lim ( \mid r \mid^n ) = 0 = \lim ( \mid r^n \mid ) = \lim ( \mid r^{ n +1} \mid )$$... and further by arguing that if $$\lim \mid r^{ n +1} \mid = 0$$ ...... then ...... $$\lim r^{ n +1} = 0$$ ...
Can someone please comment on my reasoning to resolve the problem I had with Sohrab's assertion that appeal to Exercise 2.2.7 would lead to $$( \ast )$$ following at once ...
Help will be appreciated ...

Peter
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The above post refers to Exercise 2.2.7 ... so I am providing text of the same ... as follows:View attachment 9049
Hope that helps ...

Peter

 

[StoneTemplePython]

A nice way to do this would be with cauchy sequences, but I don't know if you've gotten to them yet. The argument would be

$m \gt n \geq N$$\big \vert s_n - s_m\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(1 -r^{n+1} -(1 -r^{m+1})\big)\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(- r^{n+1} + r^{m+1}\big)\big \vert $
$ = \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\cdot \big \vert r^{m-n} - 1 \big \vert $
$ \lt 2 \cdot \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1} $

where the inequality follows from triangle inequality

select $\frac{\epsilon}{2}$
i.e. based on earlier work with the non-negative case, you know
$\big(\big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\big)$ tends to zero... so re-use this result
- - - -
as far as your suggested approach, I think it works.

You can also observe that
$\big \vert r\big \vert^n =\big \vert r^n \big \vert = \big \vert r^n - L \big \vert$
with $L =0$

since you know from that prior exercise 2.2.7 that for any $\epsilon \gt 0$
$\big \vert r\big \vert^n \lt \epsilon$
by selecting large enough $N$, then you also know by selecting that same $N$ that
$\big \vert r^n - L \big \vert \lt \epsilon$
which is your definition of a limit

 

[Math Amateur]

A nice way to do this would be with cauchy sequences, but I don't know if you've gotten to them yet. The argument would be

$m \gt n \geq N$$\big \vert s_n - s_m\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(1 -r^{n+1} -(1 -r^{m+1})\big)\big \vert $
$ = \big \vert\big(\frac{a}{1-r}\big)\big(- r^{n+1} + r^{m+1}\big)\big \vert $
$ = \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\cdot \big \vert r^{m-n} - 1 \big \vert $
$ \lt 2 \cdot \big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1} $

where the inequality follows from triangle inequality

select $\frac{\epsilon}{2}$
i.e. based on earlier work with the non-negative case, you know
$\big(\big \vert\frac{a}{1-r}\big \vert \big \vert r\big \vert^{n+1}\big)$ tends to zero... so re-use this result
- - - -
as far as your suggested approach, I think it works.

You can also observe that
$\big \vert r\big \vert^n =\big \vert r^n \big \vert = \big \vert r^n - L \big \vert$
with $L =0$

since you know from that prior exercise 2.2.7 that for any $\epsilon \gt 0$
$\big \vert r\big \vert^n \lt \epsilon$
by selecting large enough $N$, then you also know by selecting that same $N$ that
$\big \vert r^n - L \big \vert \lt \epsilon$
which is your definition of a limit

Thanks steep ...

Most grateful for your help ...

Peter