infinity = -1?

[moth]

I saw a video on youtube showing infinity = -1.

Basically it starts with the series 1+2+4+8+16+... = infinity, then multiplies each term by (2-1) which of course equals 1 and 1 * x = x so you should wind up with an identical series.

Now you have 2 series 2+4+8+16... and -1-2-4-8-16... and all the terms of the two series cancel each other except -1 so infinity = -1.

I'm suspicious of the result but I don't see any mistake(I'm no math wizz).

So I tried the same thing with 3-2=1 but I soon realized any other combination (3-2,4-3, etc)=1 just gives the original series back as you would expect multiplying by 1 to do.

Is infinity = -1?

[timo]

The limit of the series is [math] +\infty[/math] before and after the multiplication. The statement in which the limit is minus one uses an ad hoc argument instead of a proper proof. The ad hoc argument implies the assumption that you can rearrange the order of the terms in a series without changing the limit of the series. This assumption is wrong, and even if it was correct I am not sure that the series would converge to -1, since I think it would alternate.

Edited September 9, 2011 by timo

[ajb]

I think timo is essentially right, the whole thing is "sleight of hand" trick with infinite series and their convergence. It is a neat trick, but is clearly wrong as infinity is not the same as -1.

[atomthick]

There is nothing odd here. The series 1+2+4+8+16+... clearly diverges but you can rewrite it in the folowing way Sum(1 + 1/(2k)^x), k=1..inf. Taking x=-1 will get you to the sum 1+2+4+6+...

 

This function clearly diverge for x <= 1 but by analytic continuation for x < 1 you can get a more general function which has the same values as our Sum. The trick with multiplying the sum with (2-1) has enabled you to find the value of the more general function i.e. for the analytic continuation.

[khaled]

Actually, the two are different, since an operation has been done over its elements,

 

So, there's no essence in relating the sum of one to the other ...

[moth]

It's a little easier for me to see the reordering if I use letters in the series

[math]a_{1}=1, a_{n+1}=2a_{n}[/math]

[math] (2-1)*(a_1+a_2+a_3+a_4...)=((2a_1-1a_1)+(2a_2-1a_2)+(2a_3-1a_3)+...)[/math] gives the original series back. But the way it's done in the video

[math] ((-1a_1)+(2a_1-1a_2)+(2a_2-1a_3)+...)[/math] is a different series so you can't say it's equal to the original anymore.

Thanks for the help!

[DrRocket]

The limit of the series is [math] +\infty[/math] before and after the multiplication. The statement in which the limit is minus one uses an ad hoc argument instead of a proper proof. The ad hoc argument implies the assumption that you can rearrange the order of the terms in a series without changing the limit of the series. This assumption is wrong, and even if it was correct I am not sure that the series would converge to -1, since I think it would alternate.

 

The nth term doesn't even tend to 0.

 

This is a classic example of why you can't do naive term-by-term operations with infinite series and expect to reach a valid conclusion.

[little boy]

the video is wrong and misleading! They play trick!

 

The two series start at the same time.

 

That means the upper"2" doesn't start with the lower"-2", instead it start with lower "-1",

 

and then after you cancel out the upper number and lower number,

 

that will be 1+2+4+8....... (the same as the original one),

 

then you get the right answer as infinity!!

 

I think the above is the simplest way to explain. Is it helpful?

Edited September 14, 2011 by little boy

[The Light Barrier]

The nth term doesn't even tend to 0.

 

This is a classic example of why you can't do naive term-by-term operations with infinite series and expect to reach a valid conclusion.

Edited September 19, 2011 by The Light Barrier