I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

Because using the "normal"/simple definitions of summation and integers, that summation does diverge. The natural numbers are closed over addition, and yet -1/12 is not a natural number, which breaks that closure of natural numbers. So in order to make sense of this contradiction, alternative/fancier definitions of summations and numbers must be used. Specifically, p-adic numbers, which converge for large values rather than diverge. Once you're using p-adic numbers, you're not using the natural numbers and aren't restricted to the closure of natural numbers, and so can achieve 1 + 2 + 3 + ... = -1/12.

Isn't 2^-1 equivalent to 1/2, so 1/2^-1 would be 1/(1/2), which would be equivalent to 2? So 1/1^-1 + 1/2^-1 + 1/3^-1 + 1/4^-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..? != sum 1/(1/n) where n=1..?

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.

But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..? != sum 1/(1/n) where n=1..?.

I am going to do a point-by-point refutation of Conrstalks's post, and I know that can come through as rude. I just want to say I don't mean to be rude or sound upset in any way.

I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

Because using the "normal"/simple definitions of summation and integers, that summation does diverge. The natural numbers are closed over addition, and yet -1/12 is not a natural number, which breaks that closure of natural numbers. So in order to make sense of this contradiction, alternative/fancier definitions of summations and numbers must be used. Specifically, p-adic numbers, which converge for large values rather than diverge. Once you're using p-adic numbers, you're not using the natural numbers and aren't restricted to the closure of natural numbers, and so can achieve 1 + 2 + 3 + ... = -1/12.

I understand what you are saying, but it doesn't apply in this case. Even when working with p-adic numbers, an infinite series can only converge if the terms have limit 0 (it turns out in p-adic numbers this condition is also sufficient). That means that you can compute things like 1*1 + 2*2 + 3*4 + 4*8 + 5*16 + 6*32 + ... in the 2-adic numbers (I think the sum is 1). But the general term in 1 + 2 + 3 + 4 + ... does not converge to zero even using the p-adic norm.

Isn't 2^-1 equivalent to 1/2, so 1/2^-1 would be 1/(1/2), which would be equivalent to 2? So 1/1^-1 + 1/2^-1 + 1/3^-1 + 1/4^-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..? != sum 1/(1/n) where n=1..?

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.

But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..? != sum 1/(1/n) where n=1..?.

1 + 2 + 3 + 4 + ... and 1/(1/1) + 1/(1/2) + 1/(1/3) + 1/(1/4) + ... are the exact same series, because their terms are the same. Of course their sums can't be different, no matter how you define them.

Isn't 2^-1 equivalent to 1/2, so 1/2^-1 would be 1/(1/2), which would be equivalent to 2? So 1/1^-1 + 1/2^-1 + 1/3^-1 + 1/4^-1 ... = 1+2+3+4...? Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.i.e. sum [font=courier new']n[/font] where [font=courier new']n=1..?[/font] != sum [font=courier new']1/(1/n)[/font] where [font=courier new']n=1..?[/font]

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.But maybe I'm not understanding, as I can't see the contradiction/inequality in sum [font=courier new']n[/font] where [font=courier new']n=1..?[/font] != sum [font=courier new']1/(1/n)[/font] where [font=courier new']n=1..?[/font].

1 + 2 + 3 + 4 + ... and 1/(1/1) + 1/(1/2) + 1/(1/3) + 1/(1/4) + ... are the exact same series, because their terms are the same. Of course their sums can't be different, no matter how you define them.

That question of mine was in response to Mats1's post above it, where he told me that one of those sums did equal -1/12, but the other did not. Seeing they're equivalent, I was asking why this is so.

As for treating infinity as a regular algebraic variable - yeah I've seen that used to "prove" that 0=1, etc...
However, wasn't the calulus-type method before calculus to introduce, say x/? for a large part of your working (allowing you to work with infinitely small 'differentials'), but then at the end make the assumption that these terms are 0 and remove it. IIRC, this is no where near as formally robust as modern calculus, but in many situations can give you the same results. I heard there were two camps at the time, those who rejected calculus out of hand as a stupid trick that doesn't make sense, and those who were willing to treat infinity as a workable value...

Don't physicists also end up with a lot of ?'a in their algebra, and have to 'normalize' them away using different methods?

• n.b. I really don't know what I'm on about here, just a curious layman!

That question of mine was in response to Mats1's post above it, where he told me that one of those sums did equal -1/12, but the other did not. Seeing they're equivalent, I was asking why this is so.

The answer is that you are correct: Those two things are the same, and however you define one of them, you should get the same answer as for the other.

As for treating infinity as a regular algebraic variable - yeah I've seen that used to "prove" that 0=1, etc...

Those "proofs" usually include invalid manipulation. Notions of number that include infinity often will break some common features that you might expect. For instance, in ordinal numbers, 1 + omega = omega, but omega + 1 is greater than omega. In cardinal numbers, 1 + aleph_0 = aleph_0 = aleph_0 + 1, and even aleph_0 * 2 = aleph_0. However, 2^aleph_0 > aleph_0.

The use of those funky names for infinities is not capricious: One has to be very precise when dealing with infinities, or you'll end up with paradoxes all over the place.

The answer is that you are correct: Those two things are the same, and however you define one of them, you should get the same answer as for the other.

I thought however, that when presented with the naive sum 1 + 2 + 3 + 4... This will diverge. However, a value can be assigned to this series, after some manipulation, that is - 1/12, but it does not reflect the simple summation of the terms of the sequence. I'm no expert of the zeta function, but I was pretty sure this is how it goes down.

@Mats1: As I said ealier, the result has been obtained using more complicated methods based on complex analysis and not using some kind of series manipulation. When you sum an infinite number of terms you have to be very careful to what you do. Most manipulations gives in fact wrong results.

A number must fullfill some rules to be considered a number. Eg zero is the unit value of the additive operation, like 1 is the unit value of multiplication operation. Therefor

Addition:
(1) x + 0 = x = 0 +x
and
(2) x + (-x) = 0 = (-x) + x 

Now lets check infinit

Addition:
(1) inf + 0 = inf = 0 + inf

But what about this ?
(2) inf + (-inf) = 0 

If for a infinit number inf = inf + X (eg inf = 1+inf) would hold true and inf would be a valid number, then following should work
      inf + (-inf) 
= 1 + inf + (-inf) 
= 1 +     0 
= 1 
!= 0 
= inf + (-inf)

Even simpe add operations would not work in a mathematically way with such a definition of infinit.

A number must fullfill some rules to be considered a number. Eg zero is the unit value of the additive operation, like 1 is the unit value of multiplication operation. Therefor

Addition:
(1) x + 0 = x = 0 +x
and
(2) x + (-x) = 0 = (-x) + x 

Now lets check infinit

Addition:
(1) inf + 0 = inf = 0 + inf

But what about this ?
(2) inf + (-inf) = 0 

If for a infinit number inf = inf + X (eg inf = 1+inf) would hold true and inf would be a valid number, then following should work
      inf + (-inf) 
= 1 + inf + (-inf) 
= 1 +     0 
= 1 
!= 0 
= inf + (-inf)

Even simpe add operations would not work in a mathematically way with such a definition of infinit.

Check out surreal numbers, which have been mentioned in this thread before. They contain many infinite numbers, but they satisfy all the properties you mention.

It is weird no one has linked to this interesting video yet:

Vi Hart just released a video about different types of infinity. If you were interested in this thread, you'll probably enjoy it. I'm sure you'll enjoy some of her other videos too.