1 + 2 + 4 + 8 + ... = -1

Snotzenfartz

S = 1 + 2 + 4 + 8 + ...

= 1 + 2(1 + 2 + 4 + ...)

= 1 + 2S

S = 1 + 2S

S = -1

Right?

BroLo

Not really. 0 =/= 1 + 2(S), so you cannot say S = -1

Snotzenfartz

Wat? Where's the error in the reasoning?

Snotzenfartz

S = 1 + 2S

So S = -1, right?

BroLo

If S* = 1+2+4+8... (is this continued to inifinity?)

Then 1 + 2(S*) = 1+1+2+4+8...

As S is already determined, you cannot then say 0 = 1+2(S*). You can say 0 = 1+2(S) but they are two separate equations. S* (in the first equations) does not equal S.

EDIT
Sorry. I see what you mean now. Yes I guess you are correct, my mistake. I didn't realise the equation was continuing on, I misread the OP completely.

Yes you are correct, in a way.

You should edit the OP and add S to the every left side of the equation, and put a plus before the "..." to show it continues to infinity.

Gurgle

Ah, I see...

S = 1 + 2 + 4 + 8 ...
= 1 + 2(1 + 2 + 4 ...)

You can only get to the second line if S = infinity, in which case 2S = S and the rest doesn't work
If S isn't infinity, then the sequence 1+2+4+8... is finite, so you will have one less term inside the brackets on the second line. Then you can't substitute S for whats in the brackets.

BroLo

Gurgle wrote: »

Ah, I see...


You can only get to the second line if S = infinity.

I think he meant that with the "..." he should have put a "+" before it to make it clear though.

Yakuza

The sum S doesn't converge, so you're essentially saying an one aleph-null infinity is the same as another aleph-null infinity. Big deal.

Gurgle

Yakuza wrote: »

The sum S doesn't converge, so you're essentially saying an one aleph-null infinity is the same as another aleph-null infinity. Big deal.

Jaysus, why can't I remember words like 'converge' when I need them.

Eliot Rosewater

It does converge using the "p-adic norm" - i.e., if you change the way in which you measure the difference between two numbers.

If you take a rational number n and write it in the form
[LATEX]\displaystyle n=\frac{2^k r}{s}[/LATEX], where r and s are odd integers, and where k is a positive or negative integer, then the 2-adic norm of n is [LATEX]\|n\|_2 =2^{-k}[/LATEX]. The norm is a measure of how big a number is.

So, for instance [LATEX]\displaystyle \| 3 \|_2 = 1; \|4\|_2 = \frac{1}{2}; \left\|\frac{7}{12}\right\|_2 = 4[/LATEX]. It's a strange notion: 3 is a "bigger" number than 4, and both are "smaller" than 7/12!

We can show that the sum in the OP converges in the 2-adic norm to -1. We look at the size of the different between the first n terms in the sum, and the supposed limit -1:
[LATEX] \displaystyle
\| S_n - (-1) \| = \left\| \sum_{k=0}^n2^k -(-1) \right\| = \| 2^{n+1} -1 +1 \| = \| 2^{n+1} \| = \frac{1}{2^{n+1}} \rightarrow 0
[/LATEX] as [LATEX]n \rightarrow \infty [/LATEX]. So in the 2-adic norm, [LATEX]\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n 2^k =-1[/LATEX] .

Yakuza

Well shut my mouth. I'd never heard of that!
I thought once you started equating infinite divergent sums with each other, all bets were off.

edit: Who'da thunk it - Wiki already has an article on it.

Eliot Rosewater

The p-adic norm actually has some pretty nice "applications" - it's not a triviality! Naturally, the Wikipedia article doesn't mention these...but I've seen some, and they are nice.

Last year at the Irish Student Math Conference a professor from NUI Maynooth proved a theorem about binomial co-efficients using p-adic norms. Fix a rational number a. Then if the denominator of the binomial coefficient [LATEX]\displaystyle {a \choose k}= \frac{a(a-1)...(a-n+1)}{k!}[/LATEX] (for k an integer) has a given prime p in it, then a also has the same p in its denominator.

The approach was to show that the function [LATEX]\displaystyle a \mapsto {a \choose k}[/LATEX] is continuous with respect to the p-adic norm, and that hence the inverse image of some closed set was closed, and then something about the inverse image only containing numbers with primes p in their denominator. It was a nice mix of arithmetic/algebra/kind of number theory with analysis.

Snotzenfartz

If we add in negative powers as well it sums to 0 since 1/2 + 1/4 + 1/8 + ... = 1.

What we get is,

... + 1/8 + 1/4 + 1/2 + 1 + 2 + 4 + 8 + ... = 0.

All the powers of 2 add up to zero. WTF is going on? :eek:

Snotzenfartz

I just realised the same is true for any number and here's why.

Define

[latex]S = ... 1/a^2 + 1/a + 1 + a + a^2 + ...[/latex]

Now multiply through by [latex]a[/latex], to give,

[latex]aS = ... 1/a + 1 + a + a^2 + a^3 ...[/latex]

The whole series has been shifted left one step which doesn't change its value because it has no beginning or end. It's the same series that we started with, it's just viewed from a different point.

So, [latex]aS = S[/latex].

Which means [latex]S = 0.[/latex]

Splagmata

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Splagmata

1+1+1+... = -1/2

...+1+1+1 = 1/2

...1+1+1+... = 0

trufax.

Capt'n Midnight

Snotzenfartz wrote: »

S = 1 + 2 + 4 + 8 + ...

= 1 + 2(1 + 2 + 4 + ...)

Right?

You have just said
S = X

= 1 + 2(X)


Even if infinity is the same as (infinity +1) it's not the same as twice infinity and neither are the same as infinite infinities. It doesn't behave as a normal number.



You've probably seen this
http://scidiv.bellevuecollege.edu/math/infinitehotel.html

Splagmata

1 + 2 + 3 + 4 + ... = -1/12

tru story.

Greenmachine

Wow