In the proof of Bolzano Weierstrass thm. about converging subsequences.

Question

Just one line is giving me trouble:

$$(2^{-l+1} + 2^{-l} + 2^{-l-1}+ \cdots + 2^{-m+2} )\cdot M$$

$$=[(2^{-l+2} - 2^{-l+1}) + (2^{-l-1} - 2^{-l}) + \cdots +(2^{-m+3} - 2^{-m+2}) ]\cdot M$$

Hows that?

Where did you get this from? The second line seems to have the summand $\,2^{-l+2}\,$ which doesn't exist in the first line...
Yeah that's where I'm coming from. It's from Real Analysis and Foundations by Steven G. Krantz. Note that they each changed to pairs of differences... Even stranger, there appears to be the same amount of terms in the second as in the first. But how could $2^{-l+1} = (2^{-l+2} - 2^{-l+1})$ ?
@lhf, It pains me to have to bother you for more details, please forgive me, but how is that? EDIT: I'm even struggling to see how to apply it to the problem at hand... -_-

EuYu · Accepted Answer · 2012-09-25 22:27:24Z

up vote 5 down vote accepted

Its from the fact that $$2^{k+1} - 2^k = 2^k(2-1) = 2^k$$ Therefore each power of $2$ is expanded as a difference.

i.e. $$2^{-l+1} = 2^{-l+2} - 2^{-l+1}$$ and so on....

answered Sep 25 '12 at 22:27

EuYu
15.3k21339

DonAntonio · Answer 2 · 2012-09-25 22:27:47Z

up vote 2 down vote

Oh, your comment made me see! Check that for any $\,n\in\Bbb N\,$:

$$2^n-2^{n-1}=2^{n-1}\left(2-1\right)=2^{n-1}...!!$$

so $\,2^{-l+1}=2^{-l+2}-2^{-l+1}\,$ and etc.

answered Sep 25 '12 at 22:27

DonAntonio
46.8k42371

2 Answers