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

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?

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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... – DonAntonio Sep 25 '12 at 22:20
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 $2^{-l+1} = (2^{-l+2} - 2^{-l+1})$ ? – tenCupMaximum Sep 25 '12 at 22:23
$2^{n+1} = 2^{n} + 2^{n}$. – lhf Sep 25 '12 at 22:25
@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... -_- – tenCupMaximum Sep 25 '12 at 22:29
$\,2^n+2^n=2\cdot 2^n=2^{n+1}\,$ – DonAntonio Sep 25 '12 at 22:30

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....

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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.

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