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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 2.3.8 ...
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
========================================================================================
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
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
========================================================================================
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