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Topic: on 2^0 + 2^{−1} + ... + 2^{
−n} <= 2 for all n

Replies: 5   Last Post: Apr 12, 2016 9:32 AM

 Messages: [ Previous | Next ]
 Don Redmond Posts: 117 Registered: 5/5/11
Re: on 2^0 + 2^{-1} + ... + 2^{-n} <= 2 for all n
Posted: Apr 9, 2016 3:35 PM

On Saturday, April 9, 2016 at 12:52:09 PM UTC-5, Daniel Bastos wrote:
> Prove by induction
>
> 2^0 + 2^{-1} + 2^{-2} + 2^{-3} + ... + 2^{-n} <= 2
>
> for all natural n.
>
> I haven't found a proof by induction.
>
> Here's my argument: the inequality describes a process by which we start
> with 2 and take off half of what we had. Because we always take half,
> we always leave half. Hence, we'll never get to zero, hence the
> procedure never takes 2 units completely.
>
> I have also observed that
>
> 2 - 2^0 + 2^{-1} + 2^{-2} + ... + 2^{-n} = 2^-(n + 1)
>
> because if we always end up with half of what we had, that half will be
> half of what we took last --- after n halves, we took last 2^-n and half
> of that is just 1/2 2^-n = 2^-(n + 1).
>
> (*) On another exercise
>
> I asked a question in alt.algebra.help on April 6th 2016. I haven't
> seen any activity there since then. I'll appreciate your attention on
> the post ``on n^2 - 1 being divisible by 24'' whose Message-ID is
> <86oa9mtn6t.fsf@toledo.com>. If you find appropriate, please cross-post
> your follow-up here so that others can consider it as well. Thank you!

Do you really need an induction proof? If you just sum the geometric series, you end up with

(1-2^-(n+1))/(1-1/2) =(2^(n+1)-1)/(2^(n+1)-2^n =(2^(n+1)-1)/2^n =2-2^(-n) < 2.

If you really do need an induction proof, it should easily follow from the the last equality above.

Don

Date Subject Author
4/9/16 Daniel Bastos
4/9/16 Don Redmond
4/9/16 magidin@math.berkeley.edu
4/10/16 R.J.Chapman
4/10/16 Jose Carlos Santos
4/12/16 Pfsszxt@aol.com