## Recommended Posts

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?

##### Share on other sites

The limit of the series is $+\infty$ 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 by timo
##### Share on other sites

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.

##### Share on other sites

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.

##### Share on other sites

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 ...

##### Share on other sites

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

$a_{1}=1, a_{n+1}=2a_{n}$

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

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

Thanks for the help!

##### Share on other sites

The limit of the series is $+\infty$ 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.

##### Share on other sites

the video is wrong and misleading! They play trick!

The two series start at the same time.

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 by little boy
##### Share on other sites

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 by The Light Barrier

## Create an account

Register a new account